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Integral of (x³−1)/((x⁴+1)(x+1)) | JEE

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

Question
∫x3−1(x4+1)(x+1) dx\displaystyle\int \dfrac{x^{3}-1}{(x^{4}+1)(x+1)}\,dx is
A14ln⁡(1+x4)+13ln⁡(1+x3)+C\dfrac{1}{4}\ln(1+x^{4})+\dfrac{1}{3}\ln(1+x^{3})+C
B14ln⁡(1+x4)−13ln⁡(1+x3)+C\dfrac{1}{4}\ln(1+x^{4})-\dfrac{1}{3}\ln(1+x^{3})+C
C14ln⁡(1+x4)−ln⁡(1+x)+C\dfrac{1}{4}\ln(1+x^{4})-\ln(1+x)+Ccorrect
D14ln⁡(1+x4)+ln⁡(1+x)+C\dfrac{1}{4}\ln(1+x^{4})+\ln(1+x)+C
Solution
Step 1: Split the numerator so the fraction breaks into standard pieces. Add and subtract x4x^{4}:
x3−1=(x4+x3)−(x4+1)=x3(x+1)−(x4+1)x^{3}-1=(x^{4}+x^{3})-(x^{4}+1)=x^{3}(x+1)-(x^{4}+1)
 Step 2: Divide each part by the denominator (x4+1)(x+1)(x^{4}+1)(x+1):
x3−1(x4+1)(x+1)=x3(x+1)(x4+1)(x+1)−x4+1(x4+1)(x+1)=x3x4+1−1x+1\dfrac{x^{3}-1}{(x^{4}+1)(x+1)}=\dfrac{x^{3}(x+1)}{(x^{4}+1)(x+1)}-\dfrac{x^{4}+1}{(x^{4}+1)(x+1)}=\dfrac{x^{3}}{x^{4}+1}-\dfrac{1}{x+1}
 Step 3: Integrate the first piece. Since x3=14ddx(x4+1)x^{3}=\dfrac{1}{4}\dfrac{d}{dx}(x^{4}+1),
∫x3x4+1 dx=14ln⁡(x4+1)\int\dfrac{x^{3}}{x^{4}+1}\,dx=\dfrac{1}{4}\ln(x^{4}+1)
 Step 4: Integrate the second piece and combine:
∫1x+1 dx=ln⁡(x+1)  ⇒  I=14ln⁡(x4+1)−ln⁡(x+1)+C\int\dfrac{1}{x+1}\,dx=\ln(x+1)\;\Rightarrow\; I=\dfrac{1}{4}\ln(x^{4}+1)-\ln(x+1)+C
 Correct answer: (3)
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