Indefinite Integration - Calculus - Mathematics Previous Year Questions

\({{3{x^2} + 1} \over {{x^2} - 6x + 8}}\) is equal to

The value of \(\int {{e^x}({x^5} + 5{x^4} + 1)\,.\,dx}\) is

The value of \(\int {{1 \over {1 + \cos 8x}}dx}\) is

The value of \(\int {{{{x^2} + 1} \over {{x^2} - 1}}dx}\) is

Integral of \(\int {{{dx} \over {{x^2}{{[1 + {x^4}]}^{3/4}}}}}\).

\(\int {{{{2^x}} \over {\sqrt {1 - {4^x}} }}dx}\) is equal to

\(\int \frac{3^x}{\sqrt{1-9^x}} d x\) is equal to

\(\int \frac{1}{x \sqrt{a x-x^2}} d x\) is

$$\int \frac{\cos 4 x+1}{\cot x-\tan x} d x=$$

$$\int e^x\left(1+\tan x+\tan ^2 x\right) d x \text { is equal to }$$

$$\int \sqrt{\operatorname{cosec} x-1} d x=$$

\(\int x^x(1+\log x) d x\) is equal to

$$\int \frac{f^{\prime}(x)}{f(x) \log (f(x))} d x \text { is equal to }$$

$$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x \text { is equal to }$$

$$\int \log x(\log x+2) d x \text { equals to }$$

$$\int e^x\left[\frac{x^2+1}{(x+1)^2}\right] d x \quad \text { is equal to }$$

$$\text { The value of } \int \frac{d x}{\sqrt{2 x-x^2}} \text { is }$$

$$\int \frac{x}{x^4-16} d x=$$

$$\int \sqrt{x^2-4 x+2} d x=$$

If \(\int \frac{1}{\sqrt{\sin ^3 x \cos x}} d x=\frac{k}{\sqrt{\tan x}}+c\) then the value of \(k\) is