Finding the integral of $1/(x^2 +2x +5)$

Question

I am working through a Pure Maths book as a hobby. I am struggling with the second part of this problem:

Differentiate $\arctan(x + a)$ with respect to x and use your result to find

a) $\displaystyle\int \frac{dx}{x^2 + 4x +5}$

b) $\displaystyle\int \frac{dx}{x^2 + 2x +5}$

I differentiated $\arctan(x + a)$ with respect to x:

$$\frac {dy}{dx} \arctan(x + a) = \frac{1}{(x + a)^2 +1}$$

From this I was able to say:

$$\int \frac{dx}{x^2 + 4x +5} = \int \frac{dx}{(x + 2)^2 +1} \implies \int \frac{dx}{x^2 + 4x +5} = \arctan(x + 2) + c$$

But I cannot apply this technique to b) because $x^2 + 2x + 5$ won't factorise.

Answer 1

You can do it as follows: $\displaystyle x^2+2x+5=(x+1)^2+4=4\left(\left(\frac{x+1}2\right)^2+1\right)$, and therefore\begin{align}\int\frac{\mathrm dx}{x^2+2x+5}&=\frac14\int\frac{\mathrm dx}{\left(\frac{x+1}2\right)^2+1}\\&=\frac14\times2\arctan\left(\frac{x+1}2\right)+C\\&=\frac12\arctan\left(\frac{x+1}2\right)+C.\end{align}

Answer 2

But I cannot apply this technique to b) because $x^2+2x+5$ won't factorise.

I think you mean, ...because $x^2+2x+5$ doesn't have remainder $1$ when you've completed the square. Which is true. However, you can get $$\frac{1}{x^2+2x+5}$$ into the form

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

by some basic manipulation:

\begin{align} \frac{1}{(x+1)^2 + 4} \\ \\ = \frac{1}{4\left(\frac{1}{4}(x+1)^2 + 1\right)} \\ \\ = \frac{1}{4} \cdot\frac{1}{\frac{1}{4}(x+1)^2 + 1} \\ \\ = \frac{1}{4} \cdot\frac{1}{\left(\frac{1}{2}\right)^2(x+1)^2 + 1} \\ \\ = \frac{1}{4} \cdot\frac{1}{\left(\frac{1}{2}\left(x+1\right)\right)^2 + 1} \\ \\ = \frac{1}{4} \cdot\frac{1}{\left(\frac{1}{2}x+\frac{1}{2}\right)^2 + 1}\\ \\ \end{align}

Now substitute $u = \frac{1}{2}x\ $ and apply the reverse chain rule (i.e. $ dx = 2du$), and then you can apply your formula.

Answer 3

HINT:

Recall that $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}$$

If you need any more help please don't hesitate to ask.

Answer 4

Complete the square in the denominator: $$\int \frac{\mathrm dx}{x^2 + 4x +5}=\int \frac{\mathrm d(x+1)}{(x+1)^2 + 4}$$ and use the basic formula $$\int\!\frac{\mathrm du}{u^2+a^2}=\frac1a\,\arctan \frac ua.$$