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}}...

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

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

If $$\int {{{\sin 2x} \over {{{(a + b\cos x)}^2}}}dx} = \alpha \left[ {{{\log }_e}\left| {a + b\cos x} \right| + {a \over {a + b\cos x}}} \right] + c...

$$\int {{{f(x)\phi '(x) + \phi (x)f'(x)} \over {(f(x)\phi (x) + 1)\sqrt {f(x)\phi (x) - 1} }}dx = } $$

If $$\int {\cos x\log \left( {\tan {x \over 2}} \right)} dx$$ = $$\sin x\log \left( {\tan {x \over 2}} \right)$$ + f(x), then f(x) is equal to (assumi...

y = $$\int {\cos \left\{ {2{{\tan }^{ - 1}}\sqrt {{{1 - x} \over {1 + x}}} } \right\}} dx$$ is an equation of a family of

If $$\int {{2^{{2^x}}}.\,{2^x}dx} = A\,.\,{2^{{2^x}}} + C$$, then A is equal to

If $$\int {{e^{\sin x}}} .\left[ {{{x{{\cos }^3}x - \sin x} \over {{{\cos }^2}x}}} \right]dx = {e^{\sin x}}f(x) + c$$, where c is constant of integrat...

If $$\int {f(x)} \sin x\cos xdx = {1 \over {2({b^2} - {a^2})}}\log (f(x)) + c$$, where c is the constant of integration, then f(x) is equal to

$$\int {\cos (\log x)dx} $$ = F(x) + C, where C is an arbitrary constant. Here, F(x) is equal to

$$\int {{{{x^2} - 1} \over {{x^4} + 3{x^2} + 1}}dx} $$ (x > 0) is

Let I = $$\left| {\int {_{10}^{19}{{\sin x} \over {1 + {x^8}}}dx} } \right|$$. Then,