I have following integral and it should be simple, however whatever substitution I use and no matter how many times I integrate it by parts (or combine both) I never get the correct solution (or any alternative solution):
$$\int \frac{dx}{(x^2 + a^2)^2}$$
I'm looking for what is on the Wolfram|Alpha in the alternative solutions section: $$\frac{\arctan(\frac{x}{a})}{2a^3} + \frac{x}{2a^2(a^2 + x^2)}+C$$
After a linear substitution we may instead look at $$\frac{1}{(1+x^2)^2}= \frac{(1+x^2)}{(1+x^2)^2}-\frac{x^2}{(1+x^2)^2}=\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}$$ where the first term is easy. For the second term we may try integration by parts $$\int x\cdot\frac{x}{(1+x^2)^2}dx=\left[x\cdot\frac{-1}{2(1+x^2)}\right]+\frac{1}{2}\int\frac{1}{1+x^2}dx=-\frac{x}{2(1+x^2)}+\frac12\arctan x$$ Ending up with $$\int\frac{dx}{(1+x^2)^2} = \frac12\arctan x +\frac{x}{2(1+x^2)}$$
$$\int\dfrac{1}{(x^2+a^2)^2}dx$$ Hint : Does there exist a function which when differentiated gives something similar to $1\over (x^2+1)$ ?
Put $\tan^{-1} \frac{x}{a} = t$. This gives $\dfrac{1}{1+(\dfrac{x}{a})^2}.\dfrac{1}{a}.dx = dt = \dfrac{a}{a^2+{x}^2}.dx$
$$\int\dfrac{1}{(x^2+a^2)^2}dx=\int \dfrac{1}{a.( (a\tan t)^2+a^2)}dt=\int \dfrac{1}{a.( a^2\tan^2 t+a^2)}dt=\int \dfrac{1}{a^3(\tan^2 t+1)}dt=\dfrac{1}{a^3}\int \cos^2 t dt$$
which will lead to the answer.
Substitute $$x=a \hspace{3pt} \tan \theta$$
$$ dx = a\hspace{3pt} \sec^2 \theta \hspace{3pt} d\theta$$
The integral
$$ \begin{align*} \int \frac{1}{(x^2+a^2)^2} \hspace{3pt}dx &= \int \frac{\hspace{3pt}a \hspace{3pt}\sec^2 \theta }{a^4 \hspace{3pt} \sec^4 \theta}\hspace{3pt} d\theta\\ &= \frac{1}{a^3} \int \frac{1}{\sec^2 \theta}\hspace{3pt} d\theta\\ &= \frac{1}{a^3} \int \cos^2 \theta\hspace{3pt} d\theta\\ &= \frac{1}{2a^3} \int 2\hspace{3pt}\cos^2 \theta\hspace{3pt} d\theta\\ &= \frac{1}{2a^3} \int (1+\cos2 \theta)\hspace{3pt} d\theta\\ &= \frac{1}{2a^3} \theta + \frac{1}{2a^3} \frac{\sin 2\theta}{2} + \text{constant} \\ &= \frac{1}{2a^3} \tan^{-1}\frac{x}{a} + \frac{1}{2a^3} \frac{\sin 2\theta}{2} + \text{constant} \\ \tag{A}\\ \end{align*} $$
But since we substituted $x=a \hspace{3pt} \tan \theta$, which is equivalent to
$$\sin \theta = \frac{x}{\sqrt{x^2+a^2}}$$ and $$\cos \theta = \frac{a}{\sqrt{x^2+a^2}}$$
$$ \sin2\theta = 2 \sin\theta \cos\theta = \frac{2xa}{x^2+a^2}$$
The integral therefore simplifies to
$$\frac{1}{2a^3} \tan^{-1}\frac{x}{a} + \frac{1}{2a^3} \frac{ax}{x^2+a^2} + \text{constant}$$
When you see $a^2 + x^2$, think of $1+\tan^2\theta=\sec^2\theta$, and write $x=a\tan\theta$.
When you see $a^2 - x^2$, think of $1-\sin^2\theta=\cos^2\theta$, and write $x=a\sin\theta$.
When you see $x^2 - a^2$, think of $\sec^2\theta -1=\tan^2\theta$, and write $x=a\sec\theta$.
Just to expand further the idea behind AD.'s answer.
Let $n\in \mathbb{N}$, $n\geq 2$ and: $$I_n(a) = \int \frac{1}{(a^2+x^2)^n}\ \text{d} x\; ,$$ where $a>0$. Multiplying and dividing by $a^2$ the RH side and adding and subtracting $x^2$ in the integrand's numerator, you get: $$\tag{1} \begin{split} I_n(a) &= \frac{1}{a^2} \int \frac{1}{(a^2+x^2)^{n-1}}\ \text{d} x - \frac{1}{a^2}\int \frac{x^2}{(a^2+x^2)^n}\ \text{d} x\\ &= \frac{1}{a^2}\ I_{n-1}(a) -\frac{1}{a^2}\int x\ \frac{x}{(a^2+x^2)^n}\ \text{d} x\; ; \end{split}$$ the integral in the rightmost side of (1) can be evaluated by parts: $$\tag{2} \begin{split} \int x\ \frac{x}{(a^2+x^2)^n}\ \text{d} x &= -\frac{1}{2(n-1)}\ \frac{x}{(a^2+x^2)^{n-1}} + \frac{1}{2(n-1)} \int \frac{1}{(a^2+x^2)^{n-1}}\\ &= -\frac{1}{2(n-1)}\ \frac{x}{(a^2+x^2)^{n-1}} + \frac{1}{2(n-1)}\ I_{n-1}(a) \end{split}$$ therefore, plugging the righmost side of (2) into (1) you get the nice recursion formula: $$\tag{3} I_n(a) = \frac{2n-3}{a^2(2n-2)}\ I_{n-1}(a) + \frac{1}{a^2(2n-2)}\ \frac{x}{(a^2+x^2)^{n-1}}\; .$$ Using (3) recursively you get: $$ \begin{split} I_n(a) &= \frac{(2n-3)(2n-5)}{a^4 (2n-2)(2n-4)}\ I_{n-2}(a) + \frac{(2n-3)}{a^4(2n-2)(2n-4)}\ \frac{x}{(a^2+x^2)^{n-2}}\\ &\phantom{=} + \frac{1}{a^2(2n-2)}\ \frac{x}{(a^2+x^2)^{n-1}}\\ &= \frac{(2n-3)(2n-5)(2n-7)}{a^6 (2n-2)(2n-4)(2n-6)}\ I_{n-3}(a) + \frac{(2n-3)(2n-5)}{a^6 (2n-2)(2n-4)(2n-6)}\ \frac{x}{(a^2+x^2)^{n-3}}\\ &\phantom{=} + \frac{(2n-3)}{a^4(2n-2)(2n-4)}\ \frac{x}{(a^2+x^2)^{n-2}} + \frac{1}{a^2(2n-2)}\ \frac{x}{(a^2+x^2)^{n-1}}\\ &=\cdots\\ &= \frac{(2n-3)!!}{a^{2n-2} (2n-2)!!}\ I_1(a)+x\ \sum_{k=1}^{n-1} \frac{1}{a^{2(n-k)}}\ \frac{(2n-3)!!}{(2k-1)!!}\ \frac{(2k-2)!!}{(2n-2)!!}\ \frac{1}{(a^2+x^2)^k}\; . \end{split}$$ Since: $$I_1(a) = \int \frac{1}{a^2+x^2}\ \text{d} x = \frac{1}{a}\ \arctan \left( \frac{x}{a}\right)$$ finally you obtain: $$ \tag{4} \begin{split} I_n(a) &= \frac{(2n-3)!!}{a^{2n-1} (2n-2)!!}\ \arctan \left( \frac{x}{a}\right) \\ &\phantom{=} +x\ \sum_{k=1}^{n-1} \frac{1}{a^{2(n-k)}}\ \frac{(2n-3)!!}{(2k-1)!!}\ \frac{(2k-2)!!}{(2n-2)!!}\ \frac{1}{(a^2+x^2)^k} \end{split}$$ which is the elementary closed form of $I_n(a)$.
If you set $n=2$ in (4) you obtain exactly: $$I_2(a) = \frac{1}{2a^3}\ \arctan \left( \frac{x}{a}\right) + \frac{x}{2a^2 (a^2+x^2)}$$ which is the correct value of your integral.
We will use integration by parts. Assume that $a$ is positive. We first make the preliminary substitution $x=at$. Then $dx=a\,dt$, and when we go through the substitution process, we end up with $\int \frac{1}{a^3}\frac{dt}{(1+t^2)^2}$.
Let's not bother to carry the constant $\frac{1}{a^3}$ around, it can be inserted at the end. So we go after $\int \frac{dt}{(1+t^2)^2}$.
We use a little trick that has a number of uses. Let $I=\int \frac{dt}{1+t^2}$. (This is not a typo!) We recognize this integral instantly, since we know that the derivative of $\arctan t$ is $\frac{1}{1+t^2}$. But let's begin to evaluate the integral by using integration by parts.
Let $u=\frac{1}{1+t^2}$ and $dv=dt$. Then $du= -\frac{2t}{(1+t^2)^2}$ and we can take $v=t$. Thus $$I=\int \frac{dt}{1+t^2}=\frac{t}{1+t^2}- \int-\frac{2t^2\,dt}{(1+t^2)^2}.$$ Get rid of the doubled minus signs, and rewrite $2t^2$ as $2+2t^2 -2$. We end up with $$I=\frac{t}{1+t^2} +\int \frac{2(1+t^2)}{(1+t^2)^2}-\int \frac{2\,dt}{(1+t^2)^2},$$ and therefore $$I=\frac{t}{1+t^2}+2I-2\int\frac{dt}{(1+t^2)^2}.$$ Thus $$2\int\frac{dt}{(1+t^2)^2}=\frac{t}{1+t^2}+I=\frac{t}{1+t^2}+\arctan t.$$ So to find $\int \frac{dt}{(1+t^2)^2}$, divide the right-hand side by $2$. And don't forget to add the arbitrary constant $C$ of integration at the end.
Consider the function
$$\mathcal I(a)=\int_{0}^{x}\frac{\mathrm dt}{t^2+a^2}=\frac1{a}\tan^{-1}\frac{x}{a}$$
Now, the expression for $\mathcal I’(a)$ can be obtained in $2$ different ways:
$1.$ Using $\mathcal I(a)=\int_{0}^{x}\frac{\mathrm dt}{t^2+a^2}$: $$\begin{align}\mathcal I’(a)&=\frac{\mathrm d}{\mathrm da}\int_{0}^{x}\frac{\mathrm dt}{t^2+a^2}\\&=\int_0^x\frac{\partial}{\partial a}\left(\frac1{t^2+a^2}\right)\mathrm dt\tag{Leibniz integral rule}\\&=-2a\int_0^x\frac{\mathrm dt}{(t^2+a^2)^2}\end{align}$$ $2.$ Using $\mathcal I(a)=\frac1{a}\tan^{-1}\frac{x}{a}$: $$\require{cancel}\begin{align}\mathcal I’(a)&=\frac{\partial}{\partial a}\left( \frac1{a}\tan^{-1}\frac{x}{a}\right)\\&=-\frac1{a^2}\tan^{-1}\frac{x}{a}-\frac{x}{a^\cancel3}\frac{\cancel{a^2}}{x^2+a^2}\end{align}$$
Hence,
$$\int_0^x\frac{\mathrm dt}{(t^2+a^2)^2}=\frac1{2a^3} \tan^{-1}\frac{x}{a}+\frac{x}{2a^2(x^2+a^2)}$$
Now, let $f(x)$ be a function such that $\frac1{(t^2+a^2)^2}=f’(t)$. Then, by the second part of the FTC:
$$f(x)-f(0)=\frac1{2a^3}\tan^{-1}\frac{x}{a}+\frac{x}{2a^2(x^2+a^2)}$$
As infinitely many functions $f(x)$ are possible, all of which differ by a constant, $$\therefore\boxed{\int\frac{\mathrm dx}{(x^2+a^2)^2}=\frac1{2a^3} \tan^{-1}\frac{x}{a}+\frac{x}{2a^2(x^2+a^2)}+C}$$
