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 4x+82x2+8x+3\frac{4x+8}{2{x}^{2}+8x+3}?

Check out the solution below!



∫4x+82x2+8x+3 dx\int \frac{4x+8}{2{x}^{2}+8x+3} \, dx

1
Use Integration by Substitution.
Let u=2x2+8x+3u=2{x}^{2}+8x+3, du=4x+8 dxdu=4x+8 \, dx

2
Using uu and dudu above, rewrite ∫4x+82x2+8x+3 dx\int \frac{4x+8}{2{x}^{2}+8x+3} \, dx.
∫1u du\int \frac{1}{u} \, du

3
The derivative of ln⁡x\ln{x} is 1x\frac{1}{x}.
ln⁡u\ln{u}

4
Substitute u=2x2+8x+3u=2{x}^{2}+8x+3 back into the original integral.
ln⁡(2x2+8x+3)\ln{(2{x}^{2}+8x+3)}

5
Add constant.
ln⁡(2x2+8x+3)+C\ln{(2{x}^{2}+8x+3)}+C

Done
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