Indefinite IntegrationmediumFree

Integral of (p·x^(p+2q−1) − q·x^(q−1))/(x^(p+q)+1)² | JEE

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

Question
∫p xp+2q−1−q xq−1x2p+2q+2xp+q+1 dx\displaystyle\int \dfrac{p\,x^{p+2q-1}-q\,x^{q-1}}{x^{2p+2q}+2x^{p+q}+1}\,dx equals
A−xpxp+q+1+C-\dfrac{x^{p}}{x^{p+q}+1}+C
Bxqxp+q+1+C\dfrac{x^{q}}{x^{p+q}+1}+C
C−xqxp+q+1+C-\dfrac{x^{q}}{x^{p+q}+1}+Ccorrect
Dxpxp+q+1+C\dfrac{x^{p}}{x^{p+q}+1}+C
Solution
Step 1: Recognise the denominator as a perfect square:
x2p+2q+2xp+q+1=(xp+q+1)2x^{2p+2q}+2x^{p+q}+1=\big(x^{p+q}+1\big)^{2}
 Step 2: Factor x2qx^{2q} out of the squared bracket. Since xp+q+1=xq(xp+x−q)x^{p+q}+1=x^{q}\big(x^{p}+x^{-q}\big),
(xp+q+1)2=x2q(xp+x−q)2\big(x^{p+q}+1\big)^{2}=x^{2q}\big(x^{p}+x^{-q}\big)^{2}
 Step 3: Factor x2qx^{2q} out of the numerator as well:
p xp+2q−1−q xq−1=x2q(p xp−1−q x−q−1)p\,x^{p+2q-1}-q\,x^{q-1}=x^{2q}\big(p\,x^{p-1}-q\,x^{-q-1}\big)
 Step 4: Cancel x2qx^{2q}:
integrand=p xp−1−q x−q−1(xp+x−q)2\text{integrand}=\dfrac{p\,x^{p-1}-q\,x^{-q-1}}{\big(x^{p}+x^{-q}\big)^{2}}
 Step 5: Substitute t=xp+x−qt=x^{p}+x^{-q}, so dt=(p xp−1−q x−q−1) dxdt=\big(p\,x^{p-1}-q\,x^{-q-1}\big)\,dx, exactly the numerator:
I=∫dtt2=−1t+C=−1xp+x−q+CI=\int\dfrac{dt}{t^{2}}=-\dfrac{1}{t}+C=-\dfrac{1}{x^{p}+x^{-q}}+C
 Step 6: Multiply numerator and denominator by xqx^{q}:
I=−xqxp+q+1+CI=-\dfrac{x^{q}}{x^{p+q}+1}+C
 Correct answer: (3)
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