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Question

Evaluate : x3(x1)(x2+1)dx.

OR

Evaluate sin xx cos xx(x+sin x) dx.

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Solution

Let I =x3(x1)(x2+1)dxI(1+x2x+1(x1)(x21))dx=I=1dx+(x2+1(x1)(x21)x(x1)(x21))dxI=x+1x1dxx(x1)(x21)dxI=x+log|x1|I1(A)NowI1=x(x1)(x2+1)dxConsiderx(x1)(x2+1)=Ax1+2Bxx2+1+Cx2+1x=A(x2+1)+2Bx(x1)+C(x1)

On equating the coefficients of like terms on both the sides, we get : A=12,B=14,C=12

I=x+log|x1|(12dxx1142xdxx2+1+12dxx2+1)

I=x+log|x1|12log|x1|+14log|x2+1|12tan1x+C

Therefore, I=x+12log|x1|+14log|x21|12tan1x+C.

OR

Let I = sin xx cos xx(x+sin x)x+x cos xx(x+sin x)dx I=x+sin xxx cos xx(x+sin x)dx

I=(x+sin xx(x+sin x))x+x cos xx(x+sin x)dx I=1xdx1+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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