Ans:Hello student,Please find the answer to your question belowDo the long division of the integrand, we have

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Integral Calculus> (x 3 -8)(x-1)/x 2 -2x+4...

(x³-8)(x-1)/x²-2x+4

ajay , 10 Years ago

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2 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello student,
Please find the answer to your question below

$I = \int \frac{(x^3-8)(x-1)}{x^2-2x+4}dx$
Do the long division of the integrand, we have
$I = \int (x^2 - \frac{16(x-1)}{x^2-2x+4}+x-2)dx$
$I = \int (x^2+x-2)dx - \int \frac{16(x-1)}{(x-1)^2+(\sqrt{3})^{2}}dx$
$I = \frac{x^3}{3}+\frac{x^2}{2}-2x - \int \frac{16(x-1)}{(x-1)^2+(\sqrt{3})^{2}}dx$
$I = \frac{x^3}{3}+\frac{x^2}{2}-2x - 8log(x^2-2x+4)+c$

Durgesh Singh

Last Activity: 7 Years ago

hello friends,lets find out answer here,for integration of f(x)=(x^3+8)(x-1)÷(x^2-2x+4)first of all take (x^3+8) and expand it as (x+2)(x^2+4-2x) now, f(x )=(x+2)(x^2-2x+4)(x-1)/(x^2-2x+4) =(x+2)(x-1) =(x^2+x-2)now by integrating f(x) we will get =>(x^3)/3 + (x^2)/2 -2x +cso here is your ...hope it will east to understand!