JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \frac{x}{x^2-4 x+8} \mathrm{~d} x=K \log \left(x^2-4 x+8\right)+

\tan ^{-1}\left(\frac{x-2}{2}\right)+C

then the value of

K

View answer and explanation

Correct option: A

1 / 2

Option A (1/2) is correct because it is the coefficient of the logarithmic term in the antiderivative after completing the square and using substitution.

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\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