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

Integral Calculus

(x+8)(x-1)/(x²-2x+4)

ajay

11 Years agoGrade

1 Answer

Jitender Singh

11 Years ago

Ans:

Hello student,
Please find the answer to your question below

$I = \int \frac{(x+8)(x-1)}{x^2-2x+4}dx$
Do the long division of the integrand,
$I = \int (\frac {3(3x-4)}{x^2-2x+4}+1)dx$
$I = 9\int \frac {(2x-2)}{x^2-2x+4}dx-\int \frac{3}{x^2-2x+4}dx+\int dx$
$I = 18\int \frac {(x-1)}{(x-1)^2+(\sqrt{3})^{2}}dx-\int \frac{3}{x^2-2x+4}dx+\int dx$
$I = 9log(x^2-2x+4)-\int \frac{3}{x^2-2x+4}dx+\int dx$
$I = 9log(x^2-2x+4)-\sqrt{3}tan^{-1}(\frac{x-1}{\sqrt{3}})+\int dx$
$I = 9log(x^2-2x+4)-\sqrt{3}tan^{-1}(\frac{x-1}{\sqrt{3}})+ x + c$