If $I=\int \sec ^{-1} \sqrt{x} d x$, then $I$ equals

$x \mathrm{sec}^{-1} \sqrt{x}-\log (1+x)+C$

$\sec ^{-1} \sqrt{x}-\tan ^{-1} \sqrt{x}+C$

$x \sec ^{-1} \sqrt{x}+\log (1+x)+C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

I=\int \sec ^{-1} \sqrt{x} d x

, then

I

equals

View answer and explanation

Correct option: C

x \sec ^{-1} \sqrt{x}-\sqrt{x-1}+C

Option C (\

x \sec^{-1} \sqrt{x} - \sqrt{x-1} + C

is correct. The integral is solved by using substitution and integration by parts.

Practice All Chapter MCQs →

\int \sin ^5 x \cos ^4 x d x=

f(x)

\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5

f\left(\frac{4}{25}\right)

\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C

\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C}

{f}\left(\frac{1}{2}\right)=2

f(1)

{C}

\int \frac{x^7}{\left(1-x^2\right)^5} d x

\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx