# Thread: [SOLVED] 4 Integrals using substitution

1. ## [SOLVED] 4 Integrals using substitution

I am stuck on 4 problems involving a substitution to be solved.
It only says it must be solved using a substitution and no other method of solving integrals.
1)$\displaystyle \int \frac{e^x dx}{e^{2x}+2e^x+1}$. I've trying a u-sub : $\displaystyle u(x)=e^x$, also $\displaystyle u(x)=e^{\frac{x}{2}}$ and $\displaystyle u(x)=e^{2x}+2e^x+1$ but in each case I failed to solve it.
2)$\displaystyle \int ln(\cos(x))tan(x)dx$. I don't really know what sub to make here. Maybe $\displaystyle u=\cos (x)$?
3)$\displaystyle \int \frac{dx}{e^x+e^{-x}}$. Making the sub $\displaystyle u(x)=e^x$, I got $\displaystyle \int \frac{dx}{e^x+e^{-x}}=\int \frac{du}{u^2-u}$ but I don't know how to finish it!
4)$\displaystyle \int \frac{dx}{1+e^x}$. I took $\displaystyle u(x)=1+e^x$ and I got an answer very similar to the 3) : $\displaystyle \int \frac{du}{u^2+u}$ but don't know how to finish it.

2. Originally Posted by arbolis
I am stuck on 4 problems involving a substitution to be solved.
It only says it must be solved using a substitution and no other method of solving integrals.
1)$\displaystyle \int \frac{e^x dx}{e^{2x}+2e^x+1}$. I've trying a u-sub : $\displaystyle u(x)=e^x$, also $\displaystyle u(x)=e^{\frac{x}{2}}$ and $\displaystyle u(x)=e^{2x}+2e^x+1$ but in each case I failed to solve it.
2)$\displaystyle \int ln(\cos(x))tan(x)dx$. I don't really know what sub to make here. Maybe $\displaystyle u=\cos (x)$?
3)$\displaystyle \int \frac{dx}{e^x+e^{-x}}$. Making the sub $\displaystyle u(x)=e^x$, I got $\displaystyle \int \frac{dx}{e^x+e^{-x}}=\int \frac{du}{u^2-u}$ but I don't know how to finish it!
4)$\displaystyle \int \frac{dx}{1+e^x}$. I took $\displaystyle u(x)=1+e^x$ and I got an answer very similar to the 3) : $\displaystyle \int \frac{du}{u^2+u}$ but don't know how to finish it.

1) your first sub is correct the integral becomes $\displaystyle \int \frac{du}{u^2+2u+1} = \int \frac{du}{(u+1)^2}$ which is friarly easy.

2) observe $\displaystyle \frac{d}{dx} ( ln(\cos(x))) = - \tan x$

3) partial fractions $\displaystyle \frac{1}{u(u-1)} = \frac{1}{u-1} - \frac{1}{u}$

Bobak

3. follow bobak for the other two

Originally Posted by arbolis
3)$\displaystyle \int \frac{dx}{e^x+e^{-x}}$. Making the sub $\displaystyle u(x)=e^x$, I got $\displaystyle \int \frac{dx}{e^x+e^{-x}}=\int \frac{du}{u^2-u}$ but I don't know how to finish it!
actually, a substitution of $\displaystyle u = e^x$ yields $\displaystyle \int \frac 1{u^2 + 1}~du$. which you should know

4)$\displaystyle \int \frac{dx}{1+e^x}$. I took $\displaystyle u(x)=1+e^x$ and I got an answer very similar to the 3) : $\displaystyle \int \frac{du}{u^2+u}$ but don't know how to finish it.
correct. by partial fractions (or otherwise) note that $\displaystyle \frac 1{u^2 + u} = \frac 1{u(u + 1)} = \frac 1u - \frac 1{u + 1}$. now continue

4. Thanks bobak!
Yeah, I've reached
$\displaystyle \int \frac{dx}{u^2+2u+1} = \int \frac{dx}{(u+1)^2}$
but was stuck. I'll check a book for formulas then...
About the rest, I'll try it tomorrow probably and will post any help I feel to ask.

5. actually, a substitution of yields . which you should know
I didn't reach this integral which is equal to $\displaystyle arctan(u)$! I will spend some time on it.

6. Originally Posted by arbolis
Thanks bobak!
Yeah, I've reached but was stuck. I'll check a book for formulas then...
About the rest, I'll try it tomorrow probably and will post any help I feel to ask.
u + 1 is a linear function, you can integrate with it much the same way you would as with a single variable (just divide by it's derivative when done, which in this case is 1, so it makes no difference). if it helps you to see things clearer, do a secondary substitution, $\displaystyle t = u + 1$, this yields $\displaystyle \int \frac 1{t^2}~dt = \int t^{-2}~dt$

7. Thanks to both I could do them all. I want to notice (for any further reader) that for the 4)$\displaystyle \int \frac{dx}{1+e^x}$, it's worth $\displaystyle \int \frac{du}{u(u-1)}$ for the u-sub $\displaystyle u=e^x+1$. Using partial fractions as said bobak we obtain the final result which is worth $\displaystyle x-ln(e^x+1)+C$.

8. It's also worth to see that $\displaystyle \frac{1}{1+e^{x}}=\frac{1+e^{x}-e^{x}}{1+e^{x}}=1-\frac{e^{x}}{1+e^{x}},$ and by integrating those are easy, no substitution needed.