Exercise
$\int\frac{y^4-8}{y^3-2y^2}dy$
Step-by-step Solution
Learn how to solve problems step by step online. Find the integral int((y^4-8)/(y^3-2y^2))dy. Divide y^4-8 by y^3-2y^2. Resulting polynomial. Expand the integral \int\left(y+2+\frac{4y^{2}-8}{y^3-2y^2}\right)dy into 3 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int ydy results in: \frac{1}{2}y^2.
Find the integral int((y^4-8)/(y^3-2y^2))dy
Final answer to the exercise
$\frac{1}{2}y^2+2y+2\ln\left|y\right|+2\ln\left|y-2\right|+\frac{-4}{y}+C_0$