2 U-Substitution problems, Calculus Help

I have been doing these and I am not getting the answer in the back of the book. Maybe I am choosing the wrong u?

a. Integral of x^2/x^3 +1 dx

b. Integral of sin x/ 1 + cos^2 x dx

I have to do this by u substitution Thanks

I'm assuming that you meant x^2/(x^3+1) dx for a. and sinx/(1+cos^2x) dx for b. if so then....

for a. let u = x^3 + 1
next take the derivative of u
du = 3x^2

now in order to get your x^2 in your original problem to equal 3x^2, you have to multiply by 3, but pull a 1/3 outside of the integral to cancel the 3.
so now you have

1/3 integral (1/u)

the integral of 1/u is lnu

now replace u with your original value to get

1/3ln(x^3+1) + C

For b.

let u = cosx
take the derivative of u to get
du = -sinx

plug your values into the integral

integral -du/(1+u^2)

pull the negative out

-integral 1/(1+u^2)

the integral of 1/(1+u^2) is arctan(u) (which is something you should try and memorize)

plug your original value for u to get

-arctan(cosx) +C

Hope this helps!

you need to add some parentheses or people can't solve it correctly.

x^2/x^3 + 1 or x^2/(x^3 +1) are 2 completely different problem for example. I'm assuming the latter.

just set u = x^3 + 1
du = 2x^2 dx
1/2 du = x^2dx

so your problem becomes
integral of (1/2u )du = 1/2 ln (u)

replace u with ( x^3 + 1) = ln ( x^3 + 1)^(1/2)

b. sin x/ (1 + cos^2 x) dx
let u = cos^2x + 1
du = 2sinx
this will be the same answer as the previous with
integral of 1/2 u du

ln (cos^2x + 1)^(1/2)

*note: 2ln x = ln x^2

a) Integral of x^2/(x^3 +1) dx
put x^3 +1=t
dt=3x^2 dx
therefore Integral of x^2/x^3 +1 dx =1/3 log t+constant = 1/3 log(x^3+1)+constant

b) Integral of sin x/ 1 + cos^2 x dx
put cos x=t
dt= -sinx dx
therefore Integral of sin x/ 1 + cos^2 x dx = -1(tan^-(t))+constant = -tan^-(cos x)+constant