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Question

Evaluate the given integral.
5x+3x2+4x+10dx

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Solution

I=(5x+3)dxx2+4x+10

=5xdxx2+4x+10+3dxx2+4x+10

=52(2x)dxx2+4x+10+3dxx2+4x+10

=52(2x+44)dxx2+4x+10+3dxx2+4x+10

=52(2x+4)dxx2+4x+10202dx1x2+4x+10+3dxx2+4x+10

=52(2x+4)dxx2+4x+1010dxx2+4x+10+3dxx2+4x+10

=52(2x+4)dxx2+4x+107dxx2+2×2x+4+6

=52(2x+4)dxx2+4x+107dx(x+2)2+(6)2

Let t=x2+4x+10dt=(2x+4)dx

=52t12dt7logx+2(x+2)2+6+c since dxx2+a2=logx+x2+a2+c

=52t12+112+17logx+2(x+2)2+6+c

=52t12127logx+2(x+2)2+6+c

=52×2t7logx+2(x+2)2+6+c

=5x2+4x+10t7logx+2(x+2)2+6+c ........where t=x2+4x+10

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