Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The value of $\int x. \; \frac{ln(x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}} dx$ equals
Option: 1
$\sqrt{1+x^{2}}\; ln(x+\sqrt{1+x^{2}})-x+C$

Option: 2
$\frac{x}{2}.ln^{2}(x+\sqrt{1+x^{2}})-\frac{x}{\sqrt{1+x^{2}}}+C$

Option: 3
$\frac{x}{2}.ln^{2}(x+\sqrt{1+x^{2}})+\frac{x}{\sqrt{1+x^{2}}}+C$

Option: 4
$\sqrt{1+x^{2}}\; ln(x+\sqrt{1+x^{2}})+x+C$

Answers (1)

best_answer

Integration By PARTS -

Let $u$ and $v$ are two functions then

$\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx$

- wherein

Where $u$ is the Ist function $v$ is he IInd function

Rule for integration by parts -

Take Ist function $(u)$ as according I L A T E

- wherein

Where ,

I : Inverse

L : Logarithmic

A : Algebraic

T : Trignometric

E : Exponential

$\int x.\frac{ln(x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}dx$

$I=\int ln(x+\sqrt{1+x^{2}}).\frac{x}{\sqrt{1+x^{2}}}dx$

$=\sqrt{1+x^{2}} ln(x+\sqrt{1+x^{2}})-\int \frac{1}{\sqrt{1+x^{2}}}.\sqrt{1+x^{2}}dx$ (using integration by parts)

$=\sqrt{1+x^{2}} ln(x+\sqrt{1+x^{2}})-x+C$

Alternate:

Let $x+\sqrt{1+x^{2}}=e^{t} \; then$

$(\dfrac{x}{\sqrt{x^2+1}}+1)dx=e^tdt$

$I=\int x \; t\; dt=\int \frac{(e^{t}-e^{-t})t dt }{2}=\frac{1}{2}t\left ( e^{t}+e^{-1} \right )$

$-\frac{1}{2}(e^{-t}-e^{-t})+C=\sqrt{1+x^{2}}\;ln\;(x+\sqrt{1+x^{2}})-x+C$

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?

anam-khan

JEE Main 2025 session 1 official answer key out; challenge by February 6; how to calculate NTA score?

anam-khan

JEE Main 2025: NIT Raipur's CSE cut-offs drop; closing ranks in popular branches

Latest Question