IIT-JEE 2012 Paper 1 Offline | Indefinite Integrals Question 2 | Mathematics

PREVIOUS NEXT

IIT-JEE 2012 Paper 1 Offline

MCQ (Single Correct Answer)

-1

The integral $\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x$ equals (for some arbitrary constant $$K$$)

$-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$

$\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$

$-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$

$\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let $$I = \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx,\,\,J = \int {{{{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}}dx.} } $$ Then

for an arbitrary constant $$C$$, the value of $$J -I$$ equals :

$${1 \over 2}\log \left( {{{{e^{4x}} - {e^{2x}} + 1} \over {{e^{4x}} + {e^{2x}} + 1}}} \right) + C$$

$${1 \over 2}\log \left( {{{{e^{2x}} + {e^x} + 1} \over {{e^{2x}} - {e^x} + 1}}} \right) + C$$

$${1 \over 2}\log \left( {{{{e^{2x}} - {e^x} + 1} \over {{e^{2x}} + {e^x} + 1}}} \right) + C$$

$${1 \over 2}\log \left( {{{{e^{4x}} + {e^{2x}} + 1} \over {{e^{4x}} - {e^{2x}} + 1}}} \right) + C$$

IIT-JEE 2007 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let $$f(x)=\frac{x}{\left(1+x^{n}\right)^{1 / n}}$$ for $$n \geq 2$$ and $$g(x)=\underbrace{(f o f o \ldots . o f)}_{f \text { occurs } n \text { times }}(x)$$. Then $$\int x^{n-2} g(x) d x$$ equals :

$$\frac{1}{n(n-1)}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+k$$

$$\frac{1}{n(n+1)}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+k$$

$$\frac{1}{n(n-1)}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+k$$

$$\frac{1}{(n+1)}\left(1+n x^{n}\right)^{1+\frac{1}{n}}+k$$

IIT-JEE 2006

MCQ (Single Correct Answer)

-1

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$ is equal to

$\frac{\sqrt{2 x^4-2 x^2+1}}{x^2}+\mathrm{C}$

$\frac{\sqrt{2 x^4-2 x^2+1}}{x^3}+\mathrm{C}$

$\frac{\sqrt{2 x^4-2 x^2+1}}{x}+\mathrm{C}$

$\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+C$

PREVIOUS NEXT

JEE Advanced Subjects

Browse all chapters by subject