Indefinite Integration · Mathematics · COMEDK
MCQ (Single Correct Answer)
$$ \text { The value of } \int \frac{d x}{\sqrt{2 x-x^2}} \text { is } $$
$$ \int e^x\left[\frac{x^2+1}{(x+1)^2}\right] d x \quad \text { is equal to } $$
If $$\int \frac{1}{\sqrt{\sin ^3 x \cos x}} d x=\frac{k}{\sqrt{\tan x}}+c$$ then the value of $$k$$ is
$$\int \sqrt{x^2-4 x+2} d x=$$
$$ \int \frac{x}{x^4-16} d x= $$
$$ \text { The value of } \int \frac{1}{x+\sqrt{x-1}} d x \text { is } $$
$$\int \frac{x d x}{2(1+x)^{3 / 2}}$$ is equal to
$$\int \frac{4^x}{\sqrt{1-16^x}} d x$$ is equal to
$$\int x^x(1+\log x) d x$$ is equal to
$$ \int \sqrt{\operatorname{cosec} x-1} d x= $$
$$ \int e^x\left(1+\tan x+\tan ^2 x\right) d x \text { is equal to } $$
$$ \int \frac{\cos 4 x+1}{\cot x-\tan x} d x= $$
$$\int \frac{1}{x \sqrt{a x-x^2}} d x$$ is
$$\int \frac{3^x}{\sqrt{1-9^x}} d x$$ is equal to
$$\int {{{{2^x}} \over {\sqrt {1 - {4^x}} }}dx} $$ is equal to
Integral of $$\int {{{dx} \over {{x^2}{{[1 + {x^4}]}^{3/4}}}}} $$.
$${{3{x^2} + 1} \over {{x^2} - 6x + 8}}$$ is equal to
The value of $$\int {{1 \over {1 + \cos 8x}}dx} $$ is
The value of $$\int {{e^x}({x^5} + 5{x^4} + 1)\,.\,dx} $$ is
The value of $$\int {{{{x^2} + 1} \over {{x^2} - 1}}dx} $$ is