Question

1) find the integral from 1to -1 if(5sinx-2tanx+3x^5)dx

2) find the integral of x^3/(x^4+1)dx

Answer 1

∫[-1,1] 5sinx - 2tanx + 3x^5 dx

If you look carefully, all three terms are odd functions, so the whole integrand is odd. So, the integral from -1 to 1 is zero.

However, if you want to do the integration, note that since tanx = sinx/cosx = -d(cosx)/cosx, we have

-5cosx + 2log cosx + 1/2 x^6

Answer 2

for ∫x^3/(x^4+1) dx

let u = x^4+1, and so du = 4x^3 dx. So, your integral becomes

∫ 1/4 du/u

which should look familiar