Now you can integrate.
PROBLEM 5. Evaluate
Answer: Let u = x^3 − 1 and du = 3x^2 dx, and substitute.
∫
Now integrate.
ln |u| + CAnd substitute back.
ln |x^3 − 1| + CPROBLEM 6. Evaluate ∫e^5 x dx.
Answer: Let u = 5x and du = 5dx. Then du = dx. Substitute in.
∫e
u duIntegrate.
eu + CAnd substitute back.
e^5 x + CPROBLEM 7. Evaluate ∫ 23 x dx.
Answer: Let u = 3x and du = 3dx. Then du = dx. Make the substitution.