If I=∫(1+x^41-x4^32dx=1√fx+C (where, C is the constant of integration) and f2=-15)/4, then the value of 2f1/(√2) is

Correct answer: 3. Given integral is I=∫1/(x^3)+x1^x2-x^23/2dx Let 1/(x^2)-x^2=t ⇒-2/(x^3)-2xdx=dt ∴I=-1/2∫dt/(t^32) =t^(-12)+C =1/(√(1x^2-x^2))+C ∴fx=1/(x^2)-x^2 ⇒f1/(√2)=2-1/2=3/2 ⇒2f1/(√2)=3

Q9 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

If $I = \int \frac{1 + x^{4}}{{(1 - x^{4})}^{\frac{3}{2}}} d x$ $= \frac{1}{\sqrt{f (x)}} + C$ (where, $C$ is the constant of integration) and $f (2) = \frac{- 15}{4}$ , then the value of $2 f (\frac{1}{\sqrt{2}})$ is

Correct answer: 3

Solution

Given integral is

I = \int \frac{\frac{1}{x^{3}} + x}{{(\frac{1}{x^{2}} - x^{2})}^{\frac{3}{2}}} d x

Let

\frac{1}{x^{2}} - x^{2} = t

\Rightarrow (- \frac{2}{x^{3}} - 2 x) d x = d t

∴ I = - \frac{1}{2} \int \frac{d t}{t^{\frac{3}{2}}}

= t^{- \frac{1}{2}} + C

= \frac{1}{\sqrt{\frac{1}{x^{2}} - x^{2}}} + C

∴ f (x) = \frac{1}{x^{2}} - x^{2}

\Rightarrow f (\frac{1}{\sqrt{2}}) = 2 - \frac{1}{2} = \frac{3}{2}

\Rightarrow 2 f (\frac{1}{\sqrt{2}}) = 3

Hence, the correct answer is 3.

If I=∫1+x41-x432dx=1fx+C (where, C is the constant of integration) and f2=-154, then the value of 2f12 is

Solution

More Indefinite Integration Questions

If $I = \int \frac{1 + x^{4}}{{(1 - x^{4})}^{\frac{3}{2}}} d x$ $= \frac{1}{\sqrt{f (x)}} + C$ (where, $C$ is the constant of integration) and $f (2) = \frac{- 15}{4}$ , then the value of $2 f (\frac{1}{\sqrt{2}})$ is