Mathematics

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


ANSWER

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


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.
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

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

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\displaystyle =\sqrt{x^{2}+x}+\frac{1}{2}\log \left | \left ( x+\frac{1}{2} \right )+\sqrt{x^{2}+x} \right |+C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$ \frac{-1}{3\left ( 1+\tan^3x \right )}+C $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) - x + C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
$$ \int \frac{1}{1+x}\;dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer