Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$

Question

Problem:
Evaluate the following integral: \begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx \\ \end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$. \begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\ \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du = \int \cos^2{u} \, du \\ \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du = \frac{\sin(u)}{4} + \frac{u}{2} \\ \int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\ \end{eqnarray*} Now, I think I am right so far but I do not know have to get rid of the $u$ in the $\cos^2(u)$ term. Please help.
Thanks
Bob

Answer 1

Since you have $x=\tan u$, think of $u$ as an angle. Then

$$\tan u = \frac{x}{1}.$$

Draw a right triangle with legs $x$ and $1$ to demonstrate this fact (with $u$ as the angle opposite the $x$.) The hypotenuse is $\sqrt{1+x^2}.$ Now you can evaluate any trig function of $u$ that you please. E.g.,

$$\cos u = \frac{1}{\sqrt{1+x^2}}.$$

So to evaluate

$$\sin 2u = 2\sin u \cos u = 2\frac{x}{\sqrt{1+x^2}}\frac{1}{\sqrt{1+x^2}} = \frac{2x}{1+x^2}.$$

Answer 2

$$\int(1+\cos2u)du=u+\dfrac{\sin2u}2+C$$

$$\sin2u=\dfrac{2\tan u}{1+\tan^2u}=?$$

Another way: $$\int\dfrac{dx}{(x^2+1)^n}=\int\dfrac1{2x}\dfrac{2x}{(x^2+1)^n}dx$$

$$=\dfrac1{2x}\int\dfrac{2x}{(x^2+1)^n}dx-\int\left(\dfrac{d(1/2x)}{dx}\int\dfrac{2x}{(x^2+1)^n}dx\right)dx=?$$

Answer 3

I suggest that you compute your primitive like this:\begin{align}\int\frac1{(x^2+1)^2}\,\mathrm dx&=\int\frac{x^2+1}{(x^2+1)^2}\,\mathrm dx-\int\frac{x^2}{(x^2+1)^2}\,\mathrm dx\\&=\arctan(x)+\frac12\int x\frac{-2x}{(x^2+1)^2}\mathrm dx\\&=\arctan(x)+\frac12\cdot\frac{x}{x^2+1}-\frac12\int\frac1{x^2+1}\,\mathrm dx.\end{align}

Answer 4

Just to suggest another method.

$$\int \frac{1}{(x^2+1)^2} \, dx =\int \frac{x^2+1-x^2}{(x^2+1)^2} \, dx =\int \frac{1}{x^2+1} \, dx -\int \frac{x^2}{(x^2+1)^2} \, dx.$$ You can compute the first antiderivative very easily and for the second one, by parts $$\int \frac{x^2}{(x^2+1)^2} \, dx= \int x \cdot \frac{x}{(x^2+1)^2} \, dx=-\frac{1}{2}\int x\cdot \frac{\partial}{\partial x}\left(\frac{1}{x^2+1}\right) \, dx.$$ I think you can now easily finish the exercice.

Answer 5

You made a little mistake $$\int \frac{1}{(x^2+1)^2} \, dx = \int \frac{\cos{(2u)} + 1}{2} \, du = \frac{\sin(u)}{4} + \frac{u}{2} \\$$

It should be $\sin(2u)$ and not $\sin(u)$ $$\displaystyle \int \cos(2u)du=\frac {\sin(2u)}2+K$$ And $$\sin(2u)=\frac {2\tan(u)}{1+\tan^2(u)}=\frac {2x}{1+x^2}$$ $$u=\arctan(x) $$

Answer 6

$$\int{\frac 12 \cos(2u)+\frac12 \space du}=\frac12\int{\cos(2u)+1\space du}$$ $$=\frac 12 \bigg(\frac 12 \sin(2u)+u\bigg)=\frac 14\sin(2u)+\frac 12$$

You just missed that when we differentiate or integrate trigonometric functions, their arguments don't change.

Answer 7

$$\int \frac{1}{(x^2+1)^2}dx$$

Apply Trig Substitution: $x=tan(u)$

Then differentiate with respect to 'x' on both sides, then we get $dx=sec^2(u)du$

$$=\int \frac{1}{cos^2(u)(tan^2(u)+1)^2}du$$

Using the identity: $tan^2(x)=-1+sec^2(x)$

$$=\int \frac{1}{(1-1+sec^2(u))^2 cos^2(u)}du$$

Notice that $$\frac{1}{(1-1+sec^2(u))^2 cos^2(u)}=\frac{1}{sec^4(u)cos^2(u)}$$

$$=\int \frac{1}{sec^4(u)cos^2(u)}du$$

$$=\int cos^2(u)du$$

Using the identity: $cos^2(x)=\frac{1+cos(2x)}{2}$

$$=\int \frac{1+cos(2u)}{2}du$$

$$=\frac12 \int 1+cos(2u)du$$

$$=\frac12(\int 1du+\int cos(2u)du)$$

$$=\frac12(u+\frac{1}{2}sin(2u))$$

After substituting back $u=arctan(x)$

$$\int \frac{1}{(x^2+1)^2}dx=\frac12(arctan(x)+\frac{1}{2} sin(2arctan(x)))+C$$

Answer 8

$$ \begin{aligned} \int \frac{1}{\left(x^{2}+1\right)^{2}} d x &=-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{x^{2}+1}\right) \\ & \stackrel{IBP}{=} -\frac{1}{2 x\left(x^{2}+1\right)}-\frac{1}{2} \int \frac{1}{x^{2}\left(x^{2}+1\right)} d x \\ &=-\frac{1}{2 x\left(x^{2}+1\right)}-\frac{1}{2} \int\left(\frac{1}{x^{2}}-\frac{1}{x^{2}+1}\right) d x \\ &=-\frac{1}{2 x\left(x^{2}+1\right)}+\frac{1}{2 x}+\frac{1}{2} \arctan x+C \\ &=\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2} \arctan x+C \end{aligned} $$