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Question

# Evaluate : ∫x3(x−1)(x2+1)dx. OR Evaluate ∫sin x−x cos xx(x+sin x) dx.

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Solution

## Let I =∫x3(x−1)(x2+1)dx⇒I∫(1+x2−x+1(x−1)(x2−1))dx=⇒I=∫1dx+∫(x2+1(x−1)(x2−1)−x(x−1)(x2−1))dx⇒I=x+∫1x−1dx−∫x(x−1)(x2−1)dx⇒I=x+log|x−1|−I1…(A)NowI1=∫x(x−1)(x2+1)dxConsiderx(x−1)(x2+1)=Ax−1+2Bxx2+1+Cx2+1⇒x=A(x2+1)+2Bx(x−1)+C(x−1) On equating the coefficients of like terms on both the sides, we get : A=12,B=−14,C=12 ∴I=x+log|x−1|−(12∫dxx−1−14∫2xdxx2+1+12∫dxx2+1) ⇒I=x+log|x−1|−12log|x−1|+14log|x2+1|−12tan−1x+C Therefore, I=x+12log|x−1|+14−log|x2−1|−12tan−1x+C. OR Let I = ∫sin x−x cos xx(x+sin x)−x+x cos xx(x+sin x)dx ⇒I=∫x+sin x−x−x cos xx(x+sin x)dx ⇒I=∫(x+sin xx(x+sin x))−x+x cos xx(x+sin x)dx ⇒I=∫1xdx−∫1+cos xx+sin xdx Put x+sinx=t⇒(1+cos x)dx=dt in 2nd integral ∴I=log|x|−∫dtt ⇒I=log|x|−log|t|+C ∴I=log|x|−log|x+sin x|+C.

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