Question

integrate the following(x3+8)(x-1)/x2-2x+4​

Answer 1

The solution is given in the above attachment.

Answer 2

$\int \dfrac{(x^{3}+8)(x-1)}{x^{2}-2x+4}  \, dx=\dfrac{x^{3} }{3} +\dfrac{x^{2} }{2}-2x+C$

Step-by-step explanation:

We have,

$I=\int \dfrac{(x^{3}+8)(x-1)}{x^{2}-2x+4}  \, dx$

To find, $\int \dfrac{(x^{3}+8)(x-1)}{x^{2}-2x+4}  \, dx=?$

∴ $I=\int \dfrac{(x^{3}+8)(x-1)}{x^{2}-2x+4}  \, dx$

$I=\int \dfrac{(x^{3}+2^3)(x-1)}{x^{2}-2x+4} \, dx$

$I=\int \dfrac{(x+2)(x^{2}-2x+4)(x-1)}{x^{2}-2x+4} \, dx$

[ ∵ $a^{3}+b^{3} =(a+b)(a^{2}-ab+b^{2})$

$I=\int (x+2)(x-1) \, dx$

$I=\int (x^{2}-x+2) \, dx$

$I=\dfrac{x^{2+1} }{2+1} +\dfrac{x^{1+1} }{1+1}-2x+C$

Where, C is called integration constant

$I=\dfrac{x^{3} }{3} +\dfrac{x^{2} }{2}-2x+C$

Hence, $\int \dfrac{(x^{3}+8)(x-1)}{x^{2}-2x+4} \, dx=\dfrac{x^{3} }{3} +\dfrac{x^{2} }{2}-2x+C$.