`int (x^3 - 8x)/x^2 dx` Find the indefinite integral.

Wednesday, February 13, 2013

`int (x^3-8x)/x^2dx`

To solve, express the integrand as two fractions with same denominators.

`=int (x^3/x^2-(8x)/x^2)dx`

Simplify each fraction.

`=int (x - 8/x)dx`

Express it as difference of two integrals.

`=int xdx - int8/xdx`

For the first integral, apply the formula `intx^ndx= x^(n+1)/(n+1)+C` .

And for the second integral, apply the formula `int 1/xdx=ln|x|+C` .

`= int xdx - 8int1/xdx`

`=x^2/2-8ln|x|+C`

Therefore, `int (x^3-8x)/x^2dx = x^2/2-8ln|x|+C` .

Ninenotes