Saturday, April 28, 2018

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

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 .

No comments:

Post a Comment