Indefinite IntegrationmediumFree

Integral of (2x+1)/(x²+4x+1)^(3/2) | JEE

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\int \dfrac{2x+1}{(x^{2}+4x+1)^{3/2}}\,dx

equals

\dfrac{x^{3}}{\sqrt{x^{2}+4x+1}}+C

\dfrac{x}{\sqrt{x^{2}+4x+1}}+C

correct

\dfrac{x^{2}}{\sqrt{x^{2}+4x+1}}+C

\dfrac{1}{\sqrt{x^{2}+4x+1}}+C

Solution

Step 1: Divide numerator and denominator by

x^{3}

. For the numerator,

\dfrac{2x+1}{x^{3}}=\dfrac{2}{x^{2}}+\dfrac{1}{x^{3}}

. For the denominator, since

x^{2}+4x+1=x^{2}\!\left(1+\dfrac{4}{x}+\dfrac{1}{x^{2}}\right)

(x^{2}+4x+1)^{3/2}=x^{3}\!\left(1+\dfrac{4}{x}+\dfrac{1}{x^{2}}\right)^{3/2}

\Rightarrow \dfrac{2x+1}{(x^{2}+4x+1)^{3/2}}=\dfrac{\tfrac{2}{x^{2}}+\tfrac{1}{x^{3}}}{\left(1+\tfrac{4}{x}+\tfrac{1}{x^{2}}\right)^{3/2}}

Step 2: Substitute

t=1+\dfrac{4}{x}+\dfrac{1}{x^{2}}

. Then

dt=\left(-\dfrac{4}{x^{2}}-\dfrac{2}{x^{3}}\right)dx=-2\!\left(\dfrac{2}{x^{2}}+\dfrac{1}{x^{3}}\right)dx\;\Rightarrow\;\left(\dfrac{2}{x^{2}}+\dfrac{1}{x^{3}}\right)dx=-\dfrac{dt}{2}

Step 3: Rewrite and integrate:

I=\int\dfrac{-dt/2}{t^{3/2}}=-\dfrac{1}{2}\int t^{-3/2}\,dt=-\dfrac{1}{2}\cdot\dfrac{t^{-1/2}}{-1/2}=t^{-1/2}=\dfrac{1}{\sqrt{t}}

Step 4: Back-substitute

t=1+\dfrac{4}{x}+\dfrac{1}{x^{2}}=\dfrac{x^{2}+4x+1}{x^{2}}

, so

\dfrac{1}{\sqrt{t}}=\dfrac{x}{\sqrt{x^{2}+4x+1}}

I=\dfrac{x}{\sqrt{x^{2}+4x+1}}+C

Correct answer: (2)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.