One more intergral: integral of x / x^2+4x+8 dx
- Thread starter nikchic5
- Start date
It's rough to use partial fractions on this one because the denominator is not easily factorable. But, we can complete the square.
\(\displaystyle \L\\\int\frac{x}{x^{2}+4x+8}dx\)
Completing the square we get:
\(\displaystyle \L\\x^{2}+4x+8=(x+2)^{2}+4\)
Then we have:
\(\displaystyle \L\\\int\frac{x}{(x+2)^{2}+4}dx\)
Let \(\displaystyle u=x+2, \;\ du=dx, \;\ x=u-2\)
Making the substitutions, this gives:
\(\displaystyle \L\\\int\frac{u-2}{u^{2}+4}du=\int\frac{u}{u^{2}+4}du-\int\frac{2}{u^{2}+4}du\)
Now, can you finish?. I believe you have an ln and an arctan in your future.
\(\displaystyle \L\\\int\frac{x}{x^{2}+4x+8}dx\)
Completing the square we get:
\(\displaystyle \L\\x^{2}+4x+8=(x+2)^{2}+4\)
Then we have:
\(\displaystyle \L\\\int\frac{x}{(x+2)^{2}+4}dx\)
Let \(\displaystyle u=x+2, \;\ du=dx, \;\ x=u-2\)
Making the substitutions, this gives:
\(\displaystyle \L\\\int\frac{u-2}{u^{2}+4}du=\int\frac{u}{u^{2}+4}du-\int\frac{2}{u^{2}+4}du\)
Now, can you finish?. I believe you have an ln and an arctan in your future.
