Chapter 7: Integrals

Ans:

Let  $ x+4=t $ 

 $ \therefore dx=dt $ 

 $ \int{\dfrac{x}{\sqrt{x+4}}}dx=\int{\dfrac{(t-4)}{\sqrt{t}}}dt $ 

 $ =\int{\left( \sqrt{t}-\dfrac{4}{\sqrt{t}} \right)}dt $ 

 $ =\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-4\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right)+C $ 

 $ =\dfrac{2}{3}{{(t)}^{\dfrac{3}{2}}}-8{{(t)}^{\dfrac{1}{2}}}+C $ 

 $ =\dfrac{2}{3}t\cdot {{t}^{\dfrac{1}{2}}}-8{{t}^{\dfrac{1}{2}}}+C $ 

 $ =\dfrac{2}{3}{{t}^{\dfrac{1}{2}}}(t-12)+C $ 

 $ =\dfrac{2}{3}{{(x+4)}^{\dfrac{1}{2}}}(x+4-12)+C $ 

 $ =\dfrac{2}{3}\sqrt{x+4}(x-8)+C $ 

where C is an arbitrary constant.

Ans:

Let  $ {{x}^{3}}-1=t $ 

 $ \therefore 3{{x}^{2}}dx=dt $ 

 $ \Rightarrow \int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}}{{x}^{5}}dx=\int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}}{{x}^{3}}{{x}^{2}}dx $ 

 $ =\int{{{t}^{\dfrac{1}{3}}}}(t+1)\dfrac{dt}{3}\ $ \ $ =\dfrac{1}{3}\int{\left({{t}^{\dfrac{4}{3}}}+{{t}^{\dfrac{1}{3}}}\right)}dt $ 

 $ =\dfrac{1}{3}\left[ \dfrac{{{t}^{\dfrac{7}{3}}}}{\dfrac{7}{3}}+\dfrac{{{t}^{\dfrac{4}{3}}}}{\dfrac{4}{3}} \right]+C $ 

 $ =\dfrac{1}{3}\left[ \dfrac{3}{7}{{t}^{\dfrac{7}{3}}}+\dfrac{3}{4}{{t}^{\dfrac{4}{3}}} \right]+C $ 

 $ =\dfrac{1}{7}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{7}{3}}}+\dfrac{1}{4}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{4}{3}}}+C $ 

where C is an arbitrary constant.

Ans:

Let  $ 2+3{{x}^{3}}=t $ 

$ \therefore 9{{x}^{2}}dx=dt $ 

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

$ =\dfrac{1}{9}\left[ \dfrac{{{t}^{-2}}}{-2} \right]+C $ 

$ =\dfrac{-1}{18}\left( \dfrac{1}{{{t}^{2}}} \right)+C $ 

$ =\dfrac{-1}{18{{\left( 2+3{{x}^{3}} \right)}^{2}}}+C $ 

where C is an arbitrary constant.

Ans:

Let  $ \log \text{x}=\text{t} $ 

 $ \dfrac{1}{x}dx=dt $ 

 $ \Rightarrow \int{\dfrac{1}{x{{(\log x)}^{m}}}}dx=\int{\dfrac{dt}{{{(t)}^{m}}}} $ 

 $ =\dfrac{{{t}^{-m+1}}}{-m+1}+C $ 

 $ =\dfrac{{{(\log x)}^{1-m}}}{(1-m)}+C $ 

where C is an arbitrary constant.

Ans:

Let  $ 9-4{{x}^{2}}=t $ 

 $ \therefore -8xdx=dt $ 

 $ \Rightarrow \int{\dfrac{x}{9-4{{x}^{2}}}}dx=\dfrac{-1}{8}\int{\dfrac{1}{t}}dt $ 

 $ =\dfrac{-1}{8}\log |t|+C $ 

 $ =\dfrac{-1}{8}\log \left| 9-4{{x}^{2}} \right|+C $  

where C is an arbitrary constant.

Ans:

Let  $ 2x+3=t $ 

 $ \therefore 2dx=dt $ 

 $ \Rightarrow \int{{{e}^{2x+3}}}dx=\dfrac{1}{2}\int{{{e}^{t}}}dt $ 

 $ =\dfrac{1}{2}\left( {{e}^{t}} \right)+C $ 

 $ =\dfrac{1}{2}\left( {{e}^{2x+3}} \right)+C $ 

where C is an arbitrary constant.

Ans:

Let  $ {{\text{x}}^{2}}=\text{t} $ 

 $ \therefore 2\text{xdx}=\text{dt} $ 

 $ \Rightarrow \int{\dfrac{x}{{{e}^{{{x}^{2}}}}}}dx=\dfrac{1}{2}\int{\dfrac{1}{{{e}^{t}}}}dt $ 

 $ =\dfrac{1}{2}\int{{{e}^{-t}}}dt $ 

 $ =\dfrac{1}{2}\left( \dfrac{{{e}^{-t}}}{-1} \right)+C $ 

 $ =-\dfrac{1}{2}{{e}^{-{{x}^{2}}}}+C $ 

 $ =\dfrac{-1}{2{{e}^{{{x}^{2}}}}}+C $ 

 where C is an arbitrary constant.

Ans:

Let  $ {{\tan }^{-1}}x=t $ 

 $ \therefore \dfrac{1}{1+{{x}^{2}}}dx=dt $ 

 $ \Rightarrow \int{\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}}dx=\int{{{e}^{t}}}dt $ 

 $ ={{e}^{t}}+C $ 

 $ ={{e}^{{{\tan }^{-1}}x}}+C $ 

where C is an arbitrary constant.

Ans:

Dividing the given function’s numerator and denominator by  $ {{e}^{x}} $ , we obtain,

\[\dfrac{\dfrac{\left( {{e}^{2x}}-1 \right)}{{{e}^{x}}}}{\dfrac{\left( {{e}^{2x}}+1 \right)}{{{e}^{x}}}}=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}\]

Let \[{{e}^{x}}+{{e}^{-x}}=t\]

\[\left( {{e}^{x}}-{{e}^{-x}} \right)dx=dt\]

\[\Rightarrow \int{\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1}}dx=\int{\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}}dx\]

\[=\int{\dfrac{dt}{t}}\]

\[=\log |t|+C\]

\[=\log |{{e}^{x}}-{{e}^{-x}}|+C\]

where C is an arbitrary constant.

Ans:

Let  $ {{e}^{2x}}+{{e}^{-2x}}=t $ 

 $ \Rightarrow 2{{e}^{2x}}-2{{e}^{-2x}}dx=dt $ 

 $ \Rightarrow 2\left( {{e}^{2x}}-{{e}^{-2x}} \right)dx=dt $ 

 $ \Rightarrow \int{\left( \dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}} \right)}dx=\int{\dfrac{dt}{2t}} $ 

 $ =\dfrac{1}{2}\int{\dfrac{1}{t}}d $ 

 $ =\dfrac{1}{2}\log |t|+C $ 

 $ =\dfrac{1}{2}\log \left| {{e}^{2x}}+{{e}^{-2x}} \right|+C $ 

where C is an arbitrary constant.