Chapter 7: Integrals

Ans: Let \[3x + 1 = t\],

Now, differentiate both sides,

$3dx = dt $

$ dx = \dfrac{{dt}}{3} $

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 1 + 4} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}\left( {\dfrac{{dt}}{3}} \right)}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\log |t + \sqrt {{t^2} + {2^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx}  = \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {9{x^2} + 6x + 5} | + C $ 

Where C is an arbitrary constant.

Ans: 

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x + 9 - 9} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {7 + 9 - {{\left( {x + 3} \right)}^3}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {16 - {{\left( {x + 3} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}  $ 

Let, 

$ x + 3 = t $

$ dx = dt $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {t^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = {\sin ^{ - 1}}\left( {\dfrac{t}{4}} \right) + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}  = {\sin ^{ - 1}}\left( {\dfrac{{x + 3}}{4}} \right) + C $ 

Where C is an arbitrary constant.

Ans: 

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + 2} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4} + 2} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \left( {\dfrac{1}{4}} \right)} }}} dx $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}  $ 

Now, let

 $x - \dfrac{3}{2} = t $

$ dx = dt $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dt}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \log |t + \sqrt {{t^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \log |\left( {x - \dfrac{3}{2}} \right) + \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \dfrac{1}{4}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx}  = \log |\left( {x - \dfrac{3}{2}} \right) + \sqrt {{x^2} - 3x + 2} | + C $ 

Where C is an arbitrary constant.

Ans: Let \[x - \dfrac{3}{2} = t\]

Now, differentiate both sides,

\[dx = dt\]

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4}} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {8 + \dfrac{9}{4} - \left( {{x^2} - 3x + \dfrac{9}{4}} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {\left( {\dfrac{{41}}{4}} \right) - {{\left( {{x^2} - \dfrac{3}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \dfrac{3}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {t^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = {\sin ^{ - 1}}\dfrac{t}{{\dfrac{{\sqrt {41} }}{2}}} + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = {\sin ^{ - 1}}\dfrac{{2\left( {x - \dfrac{3}{2}} \right)}}{{\sqrt {41} }} + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx}  = {\sin ^{ - 1}}\dfrac{{2x - 3}}{{\sqrt {41} }} + C $ 

Where C is an arbitrary constant.

Ans: 

$ \left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + ab $

$ \left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + \dfrac{{{{\left( {a + b} \right)}^2}}}{4} - \dfrac{{{{\left( {a + b} \right)}^2}}}{4} + ab $

$ \left( {x - a} \right)\left( {x - b} \right) = {\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]^2} - \dfrac{{{{\left( {a - b} \right)}^2}}}{4} $ 

Now, let

$x - \left( {\dfrac{{a + b}}{2}} \right) = t $

$ dx = dt $ 

Now,

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]}^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \log |t + \sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} | + C $

$ \displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx}  = \log |\left\{ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right\} + \sqrt {\left( {x - a} \right)\left( {x - b} \right)} | + C $ 

Where C is an arbitrary constant.

Ans: Let \[2{x^2} + x - 3 = t\],

Now, differentiate both sides,

\[\left( {4x + 1} \right)dx = dt\]

Now,

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = \displaystyle \int {\dfrac{1}{{\sqrt t }}} dt $

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = \dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + C $

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = 2\sqrt t  + C $

$ \displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}  = 2\sqrt {2{x^2} + x - 3}  + C $ 

Where C is an arbitrary constant.

Ans: 

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}  + \displaystyle \int {\dfrac{2}{{\sqrt {{x^2} - 1} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{   }}\left[ {eq.1} \right] $ 

Now, for \[\dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx} \], let \[{x^2} - 1 = t\]

Now, differentiate both sides,

\[2xdx = dt\]

Now, in equation \[1\],

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\displaystyle \int {\dfrac{{dt}}{{\sqrt t }}}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \dfrac{1}{2}\left( {2\sqrt t } \right) + 2\log |x + \sqrt {{x^2} - 1} | + C $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx}  = \sqrt {{x^2} - 1}  + 2\log |x + \sqrt {{x^2} - 1} | + C $ 

Where C is an arbitrary constant.

Ans: 

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \displaystyle \int {\dfrac{{5x}}{{3{x^2} + 2x + 1}}dx}  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x}}{{3{x^2} + 2x + 1}}dx}  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2 - 2}}{{3{x^2} + 2x + 1}}dx}  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx - \dfrac{5}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} }  - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} {\text{   }}\left[ {eq.1} \right] $ 

Now, for\[\dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx} \]

Let \[3{x^2} + 2x + 1 = t\],

Now, differentiate both sides,

\[\left( {6x + 2} \right)dx = dt\]

Now, in eq. \[1\],

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x} \right) + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x + \dfrac{1}{9} - \dfrac{1}{9}} \right) + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} - \dfrac{1}{3} + 1}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} + \dfrac{2}{3}}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt}  - \dfrac{{11}}{9}\displaystyle \int {\dfrac{1}{{{{\left( {x + \dfrac{1}{3}} \right)}^2} + {{\left( {\dfrac{{\sqrt 2 }}{3}} \right)}^2}}}dx}  $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\log |t| - \dfrac{{11}}{9}\left( {\dfrac{3}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C $

$ \displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx}  = \dfrac{5}{6}\log |3{x^2} + 2x + 1| - \dfrac{{11}}{{3\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C $ 

Where C is an arbitrary constant.

Ans: 

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }}}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = \displaystyle \int {\dfrac{{6x}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}} dx $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9 + 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + \displaystyle \int {\dfrac{{27}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx} {\text{  }}\left[ {eq.1} \right] $ 

Now for \[3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} \],

Let \[{x^2} - 9x + 20 = t\],

Differentiate both sides,

\[\left( {2x - 9} \right)dx = dt\]

Now, from eq. \[1\],

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + \dfrac{{81}}{4} - \dfrac{{81}}{4} + 20} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt}  + 34\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\left( {2\sqrt t } \right) + 34\log |\left( {x - \dfrac{9}{2}} \right) + \sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} | + C $

$ \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx}  = 3\left( {2\sqrt {{x^2} - 9x + 20} } \right) + 34\log |\left( {x - \dfrac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} | + C $ 

Where C is an arbitrary constant.

Ans:

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  = \displaystyle \int {\dfrac{x}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{ - 2x}}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x - 4}}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 2} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx} {\text{   }}\left[ {eq.1} \right] $ 

Now, for \[ - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx} \],

Let \[4x - {x^2} = t\],

Now, differentiate both sides,

\[\left( {4 - 2x} \right)dx = dt\]

Now, from eq. \[1\],

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt { - \left( {{x^2} - 4x + 4 - 4} \right)} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}dx}  $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \dfrac{1}{2}\left( {2\sqrt t } \right) + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C $

$ \displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx}  =  - \sqrt {4x - {x^2}}  + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C $ 

Where C is an arbitrary constant.