If $I=\int_1^{\infty} \frac{d x}{x^2 \sqrt{1+x}}$, then I equals

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Medium

1 of 1

Application +4 | -1

I=\int_1^{\infty} \frac{d x}{x^2 \sqrt{1+x}}

, then I equals

View answer and explanation

Correct option: C

\log (\sqrt{2}+1)+\sqrt{2}

Option C is correct because the integral simplifies to the integral of -1 from -1 to 1, due to the odd function properties, resulting in -2.

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