Can't find integral of $ \int \frac{1}{x^2+2ax+b^2} dx $

Question

I have been struggling for the past 2 hours to solve the following integral: $$ \int \frac{1}{x^2+2ax+b^2} dx $$ I have tested some replacements for x, like $$u = \frac{ab + x}{ab}$$ and $$u = \frac{a + x}{\sqrt{b^2 - a^2}}$$ , but nothing seems to work. The solution according to Wolfram is the following: $$ \frac{\arctan(\frac{a + x}{\sqrt{b^2 - a^2}})}{\sqrt{b^2 - a^2}} $$ I have also tried to go backwards from this solution. Do you have any suggestions for replacing x. Should I use some other method to solve it? Thanks in advance.

Answer 1

HINT: Complete the square in the denominator, and then using a suitable substitution make use of the formula $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan{\frac{x}{a}}+C$$

In general, whenever you have an integral of the form $$\int\frac{1}{ax^2+bx+c}dx$$ often the easiest and quickest way to approach it is to complete the square in the denominator, make a suitable substitution and then make use of the standard integral above, or if necessary use the standard integral $$\int\frac{1}{x^2-a^2}dx=-\frac{1}{a}\operatorname{artanh}\frac{x}{a}+C$$

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I hope that helps. If you need any more help please don't hesitate to ask :)

Answer 2

$\int \frac{1}{x^2+2ax+b^2} dx$ = $\int \frac{1}{(x+a)^2+b^2-a^2} dx$

It becomes $\int \frac{1}{(x+a)^2+c^2} dx$ = $c^2$$\int \frac{1}{\frac{(x+a)^2}{c^2}+1} dx$

Now take y = (x+a)/c

So it becomes $\frac{1}{c^2}\int\frac{1}{y^2+1}$ now its in tan inverse form

Note: I have considered $b^2 - a^2$ positive. If it is negative you have to factorize it and use log form.

Answer 3

This can be done using a standard formula:

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

Proof:

In the above integral, substitute $ x = k \tan u \Rightarrow u = \arctan (\frac{x}{k}) $

Differentiating, we get $ dx = k\sec^2 u \ du $. The integral becomes:

$$ k\int \frac{\sec^2u \ du}{k^2 \tan^2 u + k^2}dx $$

Factoring out the k, taking it out of the integral and simplifying:

$$ \frac{1}{k} \int \frac{\sec^2 u}{1 + \tan^2 u} du $$

Using $1 + \tan^2\theta = \sec^2\theta$, This simplifies to $\frac{1}{k}\int du = \frac{u}{k} + C$

Recall that $u = \arctan(\frac{x}{k})$, and we get the formula that was given above.

Also note that if $\int f(x) dx = F(x) + C$, then $\int f(ax + b) dx = \frac{1}{a}F(ax +b) + C$

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Now, for your problem, we first complete the square: Since $ (x+a)^2 = x^2 + 2ax + a^2$, we add $b^2$ and subtract $a^2$. $b^2 - a^2 $ is a constant, so it fulfills the role of $k^2$ in our previous formula. Thus $k = \sqrt{b^2 - a^2}$

SO $x^2 + 2ax + b^2 = (x+a)^2 + (\sqrt{b^2 - a^2})^2$

Substitute this in the integral formula at the top and you get the answer

$$\int \frac{1}{x^2 + 2ax + b^2}dx = \frac{1}{\sqrt{b^2 - a^2}} \arctan(\frac{x+a}{\sqrt{b^2 - a^2}}) + C $$