So the first one is -2 arcsin x + C. Is arcsin(x) that same as the inverse sin, written as sin^-^1(x) ?? then the answer is -2sin^-^1(x)+C ?
For the second one, is the work I first tried to do wrong?
and when you say split the integral so it looks like: \int \frac{2x+6}{x^2+6x -3}...
I have no idea, that is why I need your help.
maybe, drop the -2 down because the integral of 1/sqrt(1-x^2) is cos(x) or sin(x). but what happens to the -2 ? please help
for the second one maybe substituting u in for the x^2 +6x-3. Then du=2x+6 so x=(1/2)du-3. and putting that in for x would...
Homework Statement Evaluate the Integrals:
\int \frac{-2}{\sqrt{1 - x^2}} dx
\int \frac{2x+5}{x^2+6x-3} dx
\int \frac{x^3}{x^2-1} dx
\int \frac{x}{1+x^4} dxVerify that this integral is correct:
\int ue^a^u du = \frac{e^a^u}{a}(u-\frac{1}{a})+C