please, help me solve int[(2x^2+x+7)/(x^2+4)^2]dx
- Thread starter spdrmncoo
- Start date
You could start by using partial fractions.
\(\displaystyle \L\\\frac{2x^{2}+x+7}{(x^{2}+4)^{2}}=\frac{2}{x^{2}+4}+\frac{x}{(x^{2}+4)^{2}}-\frac{1}{(x^{2}+4)^{2}}\)
Now, try u-substitution or trigonometric substitution.
For the \(\displaystyle \L\\\frac{1}{(x^{2}+4)^{2}}\), try using
\(\displaystyle \L\\x=2tan(u)\) and \(\displaystyle \L\\dx=\frac{2}{cos^{2}(u)}du\)
I think there's going be an arctan involved.
Can you make any headway now?.
\(\displaystyle \L\\\frac{2x^{2}+x+7}{(x^{2}+4)^{2}}=\frac{2}{x^{2}+4}+\frac{x}{(x^{2}+4)^{2}}-\frac{1}{(x^{2}+4)^{2}}\)
Now, try u-substitution or trigonometric substitution.
For the \(\displaystyle \L\\\frac{1}{(x^{2}+4)^{2}}\), try using
\(\displaystyle \L\\x=2tan(u)\) and \(\displaystyle \L\\dx=\frac{2}{cos^{2}(u)}du\)
I think there's going be an arctan involved.
Can you make any headway now?.
Hello, spdrmncoo!
An interesting problem . . . it can be done without Partial Fractions.
The numerator is: \(\displaystyle \,2x^2\,+\,x\,+\,7\;= \;8\tan^2\theta\,+\,2\tan\theta\,+\,7\)
The denominator is: \(\displaystyle \,(x^2\,+\,4)^2\;=\;(4\tan^2\theta\,+\,4)^2\;=\
4\sec^2\theta)^2\;=\;16\sec^4\theta\)
The integral becomes: \(\displaystyle \L\,\int\frac{8\tan^2\theta\,+\,2\tan\theta\,+\,7}{16\sec^4\theta}\)\(\displaystyle \cdot(2\sec^2\theta\,d\theta) \;=\;\L\frac{1}{8}\int\frac{8\tan^2\theta\,+\,2\tan\theta\,+\,7}{\sec^2\theta}\ d\theta\)
Multiply top and bottom by \(\displaystyle \cos^2\theta:\L\;\frac{1}{8}\int\left(8\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\cos^2\theta\right)\ d\theta\)
We have: \(\displaystyle \L\:\frac{1}{8}\int\left(\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\sin^2\theta\,+\,7\cos^2\theta\right)\ d\theta\)
. . . . . \(\displaystyle \L=\;\frac{1}{8}\int\left(\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\right)\ d\theta\)
Hence: \(\displaystyle \L\:\frac{1}{8}\int \left(\frac{1\,-\,\cos2\theta}{2}\,+\,\sin2\theta\,+\,7\right)\ d\theta\;\) . . . etc.
An interesting problem . . . it can be done without Partial Fractions.
Let \(\displaystyle x\,=\,2\tan\theta\;\;\Rightarrow\;\;dx\,=\,2\sec^2\theta\,d\theta\)
The numerator is: \(\displaystyle \,2x^2\,+\,x\,+\,7\;= \;8\tan^2\theta\,+\,2\tan\theta\,+\,7\)
The denominator is: \(\displaystyle \,(x^2\,+\,4)^2\;=\;(4\tan^2\theta\,+\,4)^2\;=\
4\sec^2\theta)^2\;=\;16\sec^4\theta\)The integral becomes: \(\displaystyle \L\,\int\frac{8\tan^2\theta\,+\,2\tan\theta\,+\,7}{16\sec^4\theta}\)\(\displaystyle \cdot(2\sec^2\theta\,d\theta) \;=\;\L\frac{1}{8}\int\frac{8\tan^2\theta\,+\,2\tan\theta\,+\,7}{\sec^2\theta}\ d\theta\)
Multiply top and bottom by \(\displaystyle \cos^2\theta:\L\;\frac{1}{8}\int\left(8\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\cos^2\theta\right)\ d\theta\)
We have: \(\displaystyle \L\:\frac{1}{8}\int\left(\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\sin^2\theta\,+\,7\cos^2\theta\right)\ d\theta\)
. . . . . \(\displaystyle \L=\;\frac{1}{8}\int\left(\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\right)\ d\theta\)
Hence: \(\displaystyle \L\:\frac{1}{8}\int \left(\frac{1\,-\,\cos2\theta}{2}\,+\,\sin2\theta\,+\,7\right)\ d\theta\;\) . . . etc.