Indefinite Integrals · Mathematics · WB JEE

Evaluate $$\int {{{{x^2}} \over {x(1 + {x^2})}}dx} $$

$$\int {{{dx} \over {x(x + 1)}}} $$ equals

where c is arbitrary constant.

The value of $$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + ..... + {n \over {{n^2} + {n^2}}}} \right]$$ is

The value of $$\int\limits_{ - 1}^1 {{{|x + 2|} \over {x + 2}}dx} $$ is

$$\int {{{dx} \over {\sin x + \sqrt 3 \cos x}}} $$ equals

$$\int {{{{{\sin }^{ - 1}}x} \over {\sqrt {1 - {x^2}} }}dx} $$ equal to

where c is an arbitrary constant

$$\int {{{\log \sqrt x } \over {3x}}dx} $$ is equal to

$$\int {{e^x}\left( {{2 \over x} - {2 \over {{x^2}}}} \right)dx} $$ is equal to

The value of the integral $$\int {{{dx} \over {{{({e^x} + {e^{ - x}})}^2}}}} $$ is

$$\int {\sqrt {1 + \cos x} dx} $$ is equal to

$$\int {{{{x^3}dx} \over {1 + {x^8}}} = } $$

$$\int {{{\cos 2x} \over {\cos x}}dx = } $$

$$\int {{{{{\sin }^8}x - {{\cos }^8}x} \over {1 - 2{{\sin }^2}x{{\cos }^2}x}}dx} $$

$$\int {{2^x}(f'(x) + f(x)\log 2)dx} $$ is

$$ \text { If } \int \frac{\log _e\left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\mathrm{f}(\mathrm{g}(x))+\mathrm{c} \text { then } $$

If $$I = \int {{{{x^2}dx} \over {{{(x\sin x + \cos x)}^2}}} = f(x) + \tan x + c} $$, then $$f(x)$$ is

If $$\int {{{dx} \over {(x + 1)(x - 2)(x - 3)}} = {1 \over k}{{\log }_e}\left\{ {{{|x - 3{|^3}|x + 1|} \over {{{(x - 2)}^4}}}} \right\} + c} $$, then the value of k is

$$I = \int {\cos (\ln x)dx} $$. Then I =

Let $$\int {{{{x^{{1 \over 2}}}} \over {\sqrt {1 - {x^3}} }}dx = {2 \over 3}g(f(x)) + c} $$ ; then

(c denotes constant of integration)