1. ## Integral question

These are some questions I can't solve on a practice test, if anyone can help me it'd be great.

1. $\int \frac{5}{x^2+6x+25}dx$

2. $\int \frac{3x^2+14x+19}{x^2+4x+5}dx$

For 1. I'm not sure what to do. Just integrating it normally doesn't seem right.
For 2. I used long division for it and got $3+\frac{2x+4}{x^2+4x+5}$ and have no idea what to do next.

And also this one(should be easier)
$\int arctan(x)dx$

2. for the first one u have

$\int\frac{5}{x^2+6x+25}dx=\int\frac{5}{(x+3)^2+16} dx$ so let $\tan(\theta)=x+3$ and im sure u can get it from there.

for the second one

we use the same idea used in the first one. $\frac{2x+4}{x^2+4x+5}=\frac{2(x+2)}{(x+2)^2+1}$.
once again we let $\tan(\theta)=x+2$.

3. Originally Posted by putnam120
for the first one u have

$\int\frac{5}{x^2+6x+25}dx=\int\frac{5}{(x+3)^2+16} dx$ so let $\tan(\theta)=x+3$ and im sure u can get it from there.
Actually there is a simpler way to get that integral (and a more formal ).

You should have memorized that,
$\int \frac{1}{1+x^2}dx=\tan^{-1} x+C$

Okay, now you complete the square,
$\int \frac{5}{(x+3)^2+16} dx$
Let, $u=x+3$ thus, $u'=1$
Substitution theorem,
$5 \int \frac{1}{u^2+16} du$
Divide through by 16,
$\frac{5}{16} \int \frac{1}{\frac{u^2}{16}+1} du$
Thus,
$\frac{5}{16} \int \frac{1}{\left( \frac{u}{4} \right)^2 +1} du$
Use linear substitution,
$t=u/4$ and $t'=1/4$
Thus, by substitution theorem,
$\frac{5}{16}\cdot \frac{4}{1} \int \frac{1}{t^2+1}dt$
Thus,
$\frac{5}{4}\tan^{-1} t+C$
Substitute,
$\frac{5}{4}\tan^{-1} (u/4)+C$
Thus,
$\frac{5}{4}\tan^{-1} \left( \frac{x+1}{4} \right)+C$

4. in your last step you mistakenly subed in $x+1$ for $u$

the correct expression is $\frac{5}{4}\tan^{-1}\left(\frac{x+3}{4}\right)+C$.

and what you did is basically the same think as making the substitution that i suggested, the only difference is that yours does not require any vast knowledge of trigonometry.