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So I was trying to use partial fraction decomposition on this problem, and I realized that it didn't work, as it is already in partial fraction decomposition form.

$$\int{\frac{3x+4}{(x^2+5)^2}dx}$$

Well, $\frac{3x+4}{(x^2+5)^2}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}$

$(x^2+5)^2[\frac{3x+4}{(x^2+5)^2}]=(x^2+5)^2[\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}]$

$3x+4=(Ax+B)(x^2+5)+(Cx+D)$

$3x+4=Ax^3+5Ax+Bx^2+5B+Cx+D$

$3x+4=(A)x^3+(B)x^2+(5A+C)x+(5B+D)$

So, $A=0$, $B=0$, $5A+C=3$, and $5B+D=4$

So, $A=0$, $B=0$, $C=3$, and $D=4$

$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{(0)x+(0)}{x^2+5}+\frac{(3)x+(4)}{(x^2+5)^2}}dx$

$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{3x+4}{(x^2+5)^2}}dx.$

As this clearly doesn't work, I am wondering what the integral would be. WolframAlpha tells me that the integral is as follows:

$$\int{\frac{3x+4}{(x^2+5)^2}dx}=\frac{1}{50}(\frac{5(4x-15)}{x^2+5}+4\sqrt{5}\tan^{-1}(\frac{x}{\sqrt{5}}))+c$$

I am not sure how to get here. Any help would be greatly appreciated.

Quanto
  • 97,352
moonrune
  • 43

2 Answers2

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hint

Write it as the sum $ I+J$ with

$$I=\frac 32\int \frac{2xdx}{(x^2+5)^2}$$ easy

and

$$J=\frac 45\int \frac{x^2+5-x^2}{(x^2+5)^2}dx$$ $$=K+L$$

where $ K =\frac 45\int \frac{1}{x^2+5}dx$ is easy and

$$L=-\frac 25\int x \frac{2x}{(x^2+5)^2}dx$$

which we integrate by parts.

hamam_Abdallah
  • 62,951
0

Note that

\begin{align} \left( \frac1{x^2+5}\right)’&= -\frac{2x}{(x^2+5)^2},\>\>\>\>\> \left( \frac x{x^2+5}\right)’= \frac{10}{(x^2+5)^2} - \frac{1}{x^2+5} \end{align} which can be combined to yield $$ \left( \frac {4x-15}{x^2+5}\right)’= \frac{10(3x+4)}{(x^2+5)^2} -\frac{4}{x^2+5} $$ Then, integrate both sides to obtain \begin{align} \int{\frac{3x+4}{(x^2+5)^2}dx} &= \frac {4x-15}{10(x^2+5)}+\frac2{5\sqrt5}\tan^{-1}\frac x{\sqrt5}+C \end{align}

Quanto
  • 97,352