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Question

Mathematics Question on integral

∫x3−1x3+xdx=\int\limits \frac{x^3 -1}{x^3 + x} dx =

A

x−log⁡x+log⁡(x2+1)−tan⁡−1x+cx - \log x + \log \left(x^{2} + 1\right)- \tan^{-1} x + c

B

x−log⁡x+12log⁡(x2+1)−tan⁡−1x+cx - \log x + \frac{1}{2} \log \left(x^{2} + 1\right)- \tan^{-1} x + c

C

x+log⁡x+log⁡(x2+1)−tan⁡−1x+cx + \log x + \log \left(x^{2} + 1\right)- \tan^{-1} x + c

D

x+log⁡x+12log⁡(x2+1)−tan⁡−1x+cx + \log x + \frac{1}{2} \log \left(x^{2} + 1\right)- \tan^{-1} x + c

Answer

x−log⁡x+12log⁡(x2+1)−tan⁡−1x+cx - \log x + \frac{1}{2} \log \left(x^{2} + 1\right)- \tan^{-1} x + c

Explanation

Solution

Let I=∫x3−1x3+xdx=∫(1−x+1x3+x)dxI = \int\limits \frac{x^3 -1}{x^3 + x} dx = \int\limits \left(1 - \frac{x + 1}{x^3 + x}\right) dx
=∫1dx−∫x+1x(x2+1)dx=x−∫x+1x(x2+1)= \int\limits 1dx - \int\limits \frac{x + 1}{x(x^2 +1 )} dx = x - \int\limits \frac{x + 1}{x(x^2 +1 )} ....(i)
Now x+1x(x2+1)=Ax+Bx+Cx2+1 \frac{x + 1}{x(x^2 +1 )} = \frac{A}{x} + \frac{Bx + C}{x^2 +1}
(By using partial fractions)
⇒  x+1=A(x2+1)+(Bx+C)x\Rightarrow \:\: x + 1 = A(x^2 + 1) +(Bx+ C)x
⇒  x+1=(A+B)x2+Cx+A\Rightarrow \:\: x + 1 = (A + B)x^2 + Cx + A
Comparing coefficients of x2,xx^2, x and constant,
we get
A+B=0,C=1,A=1A + B = 0, C = 1, A = 1
⇒ B=−1\Rightarrow \: B = -1
∴  \therefore \:\: From (i), we get
I=x−∫1xdx−∫1−xx2+1dxI = x - \int\limits \frac{1}{x} dx - \int\limits \frac{1 -x }{x^2 + 1} dx
=x−log⁡ x−∫1x2+1dx+12∫2xx2+1dx= x - \log \, x - \int\limits \frac{1}{x^2 + 1} dx + \frac{1}{2} \int\limits \frac{2x}{x^2 + 1} dx
=x−log⁡ x−tan⁡−1x+12log⁡(x2+1)+c= x - \log \, x - \tan^{-1} x + \frac{1}{2} \log (x^2 + 1) + c