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Question

Q. Evaluate : 6x+7x-5) (x-4)dx

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Solution

Dear Student,

I=6x+7x-5x-4dx=6x+7x2-9x+20dx=3×2x-9+34x2-9x+20dx=3×2x-9x2-9x+20dx+34x2-9x+20dx=3×2x-9x2-9x+20dx+34×1x2-9x+20dxsolving 3×2x-9x2-9x+20dxlet x2-9x+20=u, than dx=du2x-93×2x-9x2-9x+20dx=3×2x-9u×du2x-9=3×duuapply power rule undu=un+1n+1we get3×u1-121-12=3×u12=6ureplace u=x2-9x+20=6x2-9x+20now integrate 34×1x2-9x+20dx=34×1x-922-14dx=34×12x-922-14dxlet 2x-9=u, then dx=du2=34×1u22-14×du2=34×112u2-1×du2=34×1u2-1dunow this is a standard integral we know 1u2-1du=logu2-1+uhence we get34×logu2-1+ureplace u=2x-9, the above we get=34×log2x-92-1+2x-9adding both integral values we get3×2x-9x2-9x+20dx+34×1x2-9x+20dx=34×log2x-92-1+2x-9+6x2-9x+20+C=34×log2x-9-12x-9+1+2x-9+6x2-9x+20+C=34×log2x-102x-8+2x-9+6x2-9x+20+C=34×log4x-5x-4+2x-9+6x2-9x+20+C=34×log2x-5x-4+2x-9+6x2-9x+20+C
Regards,

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