$\displaystyle \int (1+e^x)^{\frac{1}{2}} dx$

I did $\displaystyle u^2 = 1+e^x$

I got $\displaystyle 2\int u^2/(u^2 -1)du$

Is this correct ?

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$\displaystyle \int x^3/(x-1)^4 dx$

I used parts $\displaystyle \int x^3(x-1)^-4 = (-1/3)(x^3/(x-1)^3 .....$

I know you could do u =x -1

But is my way wrong ?

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$\displaystyle \int 1/(x)(4x^2 +1)^{\frac{1}{2}}dx$

i did

u ^2 = (4x^2 +1)

I got the integral to

$\displaystyle \int 1/(u^2-1)du$

Is this right ?

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I just want you to tell me if i'm right because my book gives different answers when i do the integral my way.