# Math Help - Stuck on several integral problems

1. ## Stuck on several integral problems

I'll appreciate any help on evaluating the following integrals:

$\int\frac{x+1}{x^2+2x+4}dx$

$\int\frac{x}{x^2+2x-3}dx$

$\int\frac{x+1}{x^2-2x-3}dx$

Thank you in advance

2. 1. $\displaystyle \int{\frac{x+1}{x^2 + 2x + 4}\,dx} = \frac{1}{2}\int{\frac{2x+2}{x^2 + 2x + 4}\,dx}$. Now make the substitution $\displaystyle u=x^2 + 2x + 4$.

2. You will need to use a Partial Fraction decomposition.

$\displaystyle \frac{A}{x+3} + \frac{B}{x-1} = \frac{x}{(x+3)(x-1)}$

$\displaystyle \frac{A(x-1) + B(x+3)}{(x+3)(x-1)} = \frac{x}{(x+3)(x-1)}$

$\displaystyle A(x-1)+B(x+3) = x$

$\displaystyle Ax-A+Bx+3B = x$

$\displaystyle (A+B)x -A+3B= 1x + 0$.

So $\displaystyle A+B = 1$ and $\displaystyle -A+3B=0$.

Solving simultaneously gives $\displaystyle A = \frac{3}{4}, B = \frac{1}{4}$.

So $\displaystyle \int{\frac{x}{x^2+2x-3}\,dx} = \frac{3}{4}\int{\frac{1}{x+3}\,dx} + \frac{1}{4}\int{\frac{1}{x-1}\,dx}$. Go from here.

3. Note that $\displaystyle \frac{x+1}{x^2-2x - 3} = \frac{x+1}{(x - 3)(x + 1)} = \frac{1}{x-3}$.

So $\displaystyle \int{\frac{x+1}{x^2-2x-3}\,dx} = \int{\frac{1}{x-3}\,dx}$. Go from here.

3. Thank you so much for the quick response.It was very helpful.
Here is what I did. Is it correct?

Problem 1 $\displaystyle x^2+2x+4=u, \frac{du}{dx}=2x+2. Hence, dx=\frac{du}{2(x+1)}$
$\displaystyle Then
\int \frac{x+1}{u}\frac{du}{2(x+1)}=\int \frac{1}{2} \frac{1}{u}du=\frac
{1}{2}\ln u =\frac{1}{2}\ln(x^2+2x+4)$

Problem 2 $\displaystyle\int \frac{1}{4(x-1)}+\frac{3}{4(x+3)} dx$
$\displaystyle\int\frac{1}{4}\frac{1}{x_{1}}+\int\f rac{3}{4}\frac{1}{x_{2}}=\frac{\ln(x-1)}{4}+\frac{3ln(x+3)}{4}+C$
$\displaystyle\frac{1}{4}\ln(x-1) + \frac{3}{4}\ln(x+3) + C
$

Problem 3 $\displaystyle\int\frac{x+1}{(x+1)(x-3)} = \int\frac{1}{x-3} dx = \ln(x-3) + C
$

4. Originally Posted by Plato
Thx about the recommendation. The site is great. You can watch the TED video of Wolfram if you haven't done it.It's very interesting.

5. Originally Posted by Boz
Thank you so much for the quick response.It was very helpful.
Here is what I did. Is it correct?

Problem 1 $\displaystyle x^2+2x+4=u, \frac{du}{dx}=2x+2. Hence, dx=\frac{du}{2(x+1)}$
$\displaystyle Then
\int \frac{x+1}{u}\frac{du}{2(x+1)}=\int \frac{1}{2} \frac{1}{u}du=\frac
{1}{2}\ln u =\frac{1}{2}\ln(x^2+2x+4)$

Problem 2 $\displaystyle\int \frac{1}{4(x-1)}+\frac{3}{4(x+3)} dx$
$\displaystyle\int\frac{1}{4}\frac{1}{x_{1}}+\int\f rac{3}{4}\frac{1}{x_{2}}=\frac{\ln(x-1)}{4}+\frac{3ln(x+3)}{4}+C$
$\displaystyle\frac{1}{4}\ln(x-1) + \frac{3}{4}\ln(x+3) + C
$

Problem 3 $\displaystyle\int\frac{x+1}{(x+1)(x-3)} = \int\frac{1}{x-3} dx = \ln(x-3) + C
$
1. Correct.

2. I don't know why you have decided midway to introduce $\displaystyle x_1$ and $\displaystyle x_2$, it's still a function of $\displaystyle x$.

Also, recall that $\displaystyle \int{\frac{1}{ax+b}\,dx} = \frac{1}{a}\ln{|ax+b|}$. You can't leave off the modulus. You could in question 1 because $\displaystyle x^2 + 2x + 4$ is always positive...

3. Again, you can't leave off the modulus.

6. Originally Posted by Prove It
1. Correct.

2. I don't know why you have decided midway to introduce $\displaystyle x_1$ and $\displaystyle x_2$, it's still a function of $\displaystyle x$.

Also, recall that $\displaystyle \int{\frac{1}{ax+b}\,dx} = \frac{1}{a}\ln{|ax+b|}$. You can't leave off the modulus. You could in question 1 because $\displaystyle x^2 + 2x + 4$ is always positive...

3. Again, you can't leave off the modulus.
I put $x_{1}$ just to make it easier for myself because I am still learning this integration technique.
So, the answers are going to be :

Pr.2 $\frac{1}{4}ln|x-1|+\frac{3}{4}ln|x+3|+C$
Pr.3 $ln|x-3|+C$ ??

7. The answers you have written for No. 2 and No. 3 are correct!