Problem of the Week

Updated at Dec 2, 2013 12:09 PM

To get more practice in calculus, we brought you this problem of the week:

How can we find the integral of \(\frac{4x+8}{2{x}^{2}+8x+3}\)?

Check out the solution below!



\[\int \frac{4x+8}{2{x}^{2}+8x+3} \, dx\]

1
Use Integration by Substitution.
Let \(u=2{x}^{2}+8x+3\), \(du=4x+8 \, dx\)

2
Using \(u\) and \(du\) above, rewrite \(\int \frac{4x+8}{2{x}^{2}+8x+3} \, dx\).
\[\int \frac{1}{u} \, du\]

3
The derivative of \(\ln{x}\) is \(\frac{1}{x}\).
\[\ln{u}\]

4
Substitute \(u=2{x}^{2}+8x+3\) back into the original integral.
\[\ln{(2{x}^{2}+8x+3)}\]

5
Add constant.
\[\ln{(2{x}^{2}+8x+3)}+C\]

Done
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