I'll appreciate any help on evaluating the following integrals:
$\displaystyle \int\frac{x+1}{x^2+2x+4}dx$
$\displaystyle \int\frac{x}{x^2+2x-3}dx$
$\displaystyle \int\frac{x+1}{x^2-2x-3}dx$
Thank you in advance
1. $\displaystyle \displaystyle \int{\frac{x+1}{x^2 + 2x + 4}\,dx} = \frac{1}{2}\int{\frac{2x+2}{x^2 + 2x + 4}\,dx}$. Now make the substitution $\displaystyle \displaystyle u=x^2 + 2x + 4$.
2. You will need to use a Partial Fraction decomposition.
$\displaystyle \displaystyle \frac{A}{x+3} + \frac{B}{x-1} = \frac{x}{(x+3)(x-1)}$
$\displaystyle \displaystyle \frac{A(x-1) + B(x+3)}{(x+3)(x-1)} = \frac{x}{(x+3)(x-1)}$
$\displaystyle \displaystyle A(x-1)+B(x+3) = x$
$\displaystyle \displaystyle Ax-A+Bx+3B = x$
$\displaystyle \displaystyle (A+B)x -A+3B= 1x + 0$.
So $\displaystyle \displaystyle A+B = 1$ and $\displaystyle \displaystyle -A+3B=0$.
Solving simultaneously gives $\displaystyle \displaystyle A = \frac{3}{4}, B = \frac{1}{4}$.
So $\displaystyle \displaystyle \int{\frac{x}{x^2+2x-3}\,dx} = \frac{3}{4}\int{\frac{1}{x+3}\,dx} + \frac{1}{4}\int{\frac{1}{x-1}\,dx}$. Go from here.
3. Note that $\displaystyle \displaystyle \frac{x+1}{x^2-2x - 3} = \frac{x+1}{(x - 3)(x + 1)} = \frac{1}{x-3}$.
So $\displaystyle \displaystyle \int{\frac{x+1}{x^2-2x-3}\,dx} = \int{\frac{1}{x-3}\,dx}$. Go from here.
Thank you so much for the quick response.It was very helpful.
Here is what I did. Is it correct?
Problem 1$\displaystyle \displaystyle x^2+2x+4=u, \frac{du}{dx}=2x+2. Hence, dx=\frac{du}{2(x+1)} $
$\displaystyle \displaystyle Then
\int \frac{x+1}{u}\frac{du}{2(x+1)}=\int \frac{1}{2} \frac{1}{u}du=\frac
{1}{2}\ln u =\frac{1}{2}\ln(x^2+2x+4)$
Problem 2$\displaystyle \displaystyle\int \frac{1}{4(x-1)}+\frac{3}{4(x+3)} dx$
$\displaystyle \displaystyle\int\frac{1}{4}\frac{1}{x_{1}}+\int\f rac{3}{4}\frac{1}{x_{2}}=\frac{\ln(x-1)}{4}+\frac{3ln(x+3)}{4}+C$
$\displaystyle \displaystyle\frac{1}{4}\ln(x-1) + \frac{3}{4}\ln(x+3) + C
$
Problem 3 $\displaystyle \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 $\displaystyle \displaystyle x_1$ and $\displaystyle \displaystyle x_2$, it's still a function of $\displaystyle \displaystyle x$.
Also, recall that $\displaystyle \displaystyle \int{\frac{1}{ax+b}\,dx} = \frac{1}{a}\ln{|ax+b|}$. You can't leave off the modulus. You could in question 1 because $\displaystyle \displaystyle x^2 + 2x + 4$ is always positive...
3. Again, you can't leave off the modulus.