Indefinite integral ∫ (1-sinx) of 1dx =?

Indefinite integral ∫ (1-sinx) of 1dx =?


∫1/(1-sinx)dx
=∫(1+sinx)/[(1-sinx)(1+sinx)]dx
=∫(1+sinx)/cos²xdx
=∫1/cos²xdx-∫d(cosx)/cos²x
=tanx+1/cosx+c



Definite integral (Xe ^ x) / (1 + x) ^ 2
e/2-1


Integration by parts
∫(0~1) xe^x/(1+x)^2 dx
=-∫(0~1) xe^x d[1/(1+x)]
=-e/2+∫(0~1) [1/(1+x)×(x+1)e^x] dx
=-e/2+∫(0~1) e^x dx
=-e/2+e-1
=e/2-1



Calculate the integral ∫ (x + 1) / (x ^ 2 + 4x + 13) DX


∫(x^3+4x^2+13x+x^2+4x+13)dx
∫(x^3+5x^2+17x+13)dx
x^4/4+5x^3/3+17x^2/2+13x