JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \sqrt{4 x^2-4 x+3} d x=\frac{1}{2} \log _e\left(2 x-1+\sqrt{4 x^2-4 x+3}\right)+f(x)+c

and then

f(x)

equals

View answer and explanation

Correct option: A

\frac{2 x-1}{4} \sqrt{4 x^2-4 x+3}

Option A is correct because the integral is solved by completing the square and using a standard integral formula, resulting in f(x) = (2x-1)/4 * sqrt(4x^2-4x+3).

More Questions from INDEFINITE INTEGRATION

Practice All Chapter MCQs →

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

f(x)

is differentibale and

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

then

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

equals If

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

is equal to If

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

and

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

, then

f(1)

is equal to (where

{C}

is an arbitrary constant)

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

is equal to:

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

equals