2

I am trying to integrate the following integral:

$$\int_{0}^{1} \frac{dx}{(x^2+1)^2}$$

However, I am always getting the wrong answer in the end.

What I have tried doing is the following: $$\int \frac{dx}{(x^2+1)^2} = \int \frac{1}{(\tan^2\alpha + 1)}\times \sec^2\alpha \text{ }d\alpha = \int \frac{1}{\sec^2\alpha}d\alpha =\int \cos^2\alpha \text{ }d\alpha$$

$$= \frac{1}{2} \int1+\cos(2\alpha) \text{ }d\alpha$$ Now, at this step I was trying to get $\alpha$ in terms of $x$ using the triangle. So I know that $\alpha = \arctan(x)$ and $\sin(2\alpha)=2\sin \alpha \cos\alpha = 2(\frac{x}{x^2+1})(\frac{1}{x^2+1})$. Upon substituting it back, I get that $$\int \frac{dx}{(x^2+1)^2} =\frac{1}{2}(\arctan(x)+\frac{x}{2(x^2+1)^2})$$ And when I try to find $\int_{0}^{1} \frac{dx}{(x^2+1)^2}$, I get: $$\int_{0}^{1} \frac{dx}{(x^2+1)^2} = \frac{1}{2}(\arctan(1)+\frac{1}{2(1^2+1)^2}) - \frac{1}{2}(\arctan(0)+\frac{0}{2(0^2+1)^2}) = 0.5177$$ However, the answer in the textbook is said to be $0.6427$ and interestingly enough, when I add a $2$ like so $$\int \frac{dx}{(x^2+1)^2} =\frac{1}{2}(\arctan(x)+\mathbf{2}\frac{x}{(x^2+1)^2})$$ I get the correct solution, which is quite odd since I based my solution on the fact that $$\frac{1}{2}\sin(2\alpha) = \cos(\alpha)\sin(\alpha)$$ So there should not be a $2$ there. I am probably missing something, which sadly I can't find...

Masacroso
  • 30,417
Pieraoh
  • 33
  • MathJax hint: if you put a backslash before common functions you get the right font and spacing, so \tan x gives $\tan x$ compared with tan x which gives $tan x$ You can also use \left( and \right) to get parentheses that grow to enclose what is in them, so \left(\frac 12 \right) gives $\left( \frac 12 \right)$ – Ross Millikan Jul 01 '20 at 02:32
  • @Bumblebee Could you please be a little more specific? All the steps that I have up until where I rewrite $\alpha$ in terms of $x$ seem to follow the solution in the textbook... – Pieraoh Jul 01 '20 at 02:33
  • @RossMillikan Thank you, I'll keep that in mind! – Pieraoh Jul 01 '20 at 02:33
  • 2
    @Pieraoh: You would need a square in the denominator. However it does not affect to the rest of the computations. Also if I am not mistaken, the indefinite integral should be $\frac{1}{2}\left(\arctan(x)+\frac{x}{(x^2+1)}\right)+C.$ Because $\sin 2\alpha=\dfrac{2\tan\alpha}{1+\tan^2\alpha}.$ – Bumblebee Jul 01 '20 at 02:37
  • Got it, thank you, everyone! – Pieraoh Jul 01 '20 at 02:42
  • 1
    Just note that $\frac{d}{d\alpha }\sin(2 \alpha )=2 \cos (2\alpha )$, therefore

    $$ \frac1{2}\int \cos (2 \alpha )\mathop{}!d \alpha =\frac1{4}\sin (2\alpha )+C $$

     – Masacroso Jul 01 '20 at 02:46

0 Answers0