Mathematics

# Evaluate $\displaystyle \int \frac{\log\left ( x+\sqrt{1+x^{2}} \right )}{\sqrt{1+x^{2}}}dx.$

$\displaystyle \frac{\left[\log\left ( x+\sqrt{1+x^{2}} \right )\right]^2}{2}+c$

##### SOLUTION
$\displaystyle I = \int \frac{\log\left ( x+\sqrt{1+x^{2}} \right )}{\sqrt{1+x^{2}}}dx.$

Put  $\displaystyle \log\left ( x+\sqrt{1+x^{2}} \right )=t\: And \frac{1}{\sqrt{1+x^{2}}}dx=dt.$

$\displaystyle \therefore I=\int tdt = \frac{t^2}{2}+c$

$I=\displaystyle \frac{\left[\log\left ( x+\sqrt{1+x^{2}} \right )\right]^2}{2}+c$

Hence option $'A'$ is the answer.

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Medium
Find
$\displaystyle \int \dfrac{x^{3}}{x-2}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate $\displaystyle \int \sqrt{\frac{1+x}{x}}dx.$
• A. $\displaystyle =\sqrt{x^{2}-x}+\frac{1}{2}\log \left | \left ( x+\frac{1}{2} \right )+\sqrt{x^{2}+x} \right |+C$
• B. $\displaystyle =\sqrt{x^{2}-x}+\frac{1}{2}\log \left | \left ( x-\frac{1}{2} \right )+\sqrt{x^{2}+x} \right |+C$
• C. $\displaystyle =\sqrt{x^{2}+x}+\frac{1}{2}\log \left ( \left ( x+\frac{1}{2} \right )+\sqrt{x^{2}+x} \right )+C$
• D. $\displaystyle =\sqrt{x^{2}+x}+\frac{1}{2}\log \left | \left ( x+\frac{1}{2} \right )+\sqrt{x^{2}+x} \right |+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The integral $\int \frac{\sin^2x\cos^2x}{\left ( \sin^5x+\cos^3x\sin^2x+\sin^3x\cos^2x+\cos^5x \right )^2}dx$ is equal to
• A. $\frac{1}{3\left ( 1+\cot^3x \right )}+C$
• B. $\frac{-1}{3\left ( 1+\cot^3x \right )}+C$
• C. $\frac{1}{3\left ( 1+\tan^3x \right )}+C$
• D. $\frac{-1}{3\left ( 1+\tan^3x \right )}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle I = \int \frac {\sec^{-1} \sqrt x - cosec^{-1} \sqrt x}{\sec^{-1} \sqrt x + cosec^{-1} \sqrt x} dx$, then I equals
• A. $\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) + x + C$
• B. $\displaystyle (4/\pi) (x \sec^{-1} \sqrt x + \sqrt {x - 1}) - x + C$
• C. none of these
• D. $\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) - x + C$

$\int \frac{1}{1+x}\;dx$