How to evaluate the following integral, $\int\frac{x \, dx}{x^2+2x+17}$?

Question

I am new to integration. This function is kinda tricky for me : $$\int\frac{x \, dx}{x^2+2x+17}$$

I came up with following three approaches:

1. Partial fraction decomposition, but I can't factor the denominator into different parts.
2. Substitution: I tried $u = x^{2}$ and $u = x^2+2x+17$, but both of them seem not to be helpful.
3. I also tried dividing both denominator and numerator by $x$, then the fraction became $\frac{1}{x+\frac{17}{x}+2}$ . It is more complex.

Any suggestions or hints? Thank you so much!

Answer 1

Variations on this one come up here from time to time.

There is a standard technique in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial without a first-degree term. $$ u = x^2+2x+17,\qquad du = (2x+2)\,dx, \qquad \frac{du} 2 = (x+1)\,dx $$ $$ \frac{x}{x^2+2x+17} = \underbrace{\frac{x+1}{x^2+2x+17}} + \frac{-1}{x^2+2x+17} $$ The integral of the function over the $\underbrace{\text{underbrace}}$ is done via the substitution above. Then next term has to be treated differently.

$$ \underbrace{x^2+2x+17 = (x^2 + 2x + 1) + 16 = (x+1)^2 + 16}_\text{completing the square} $$

$$ \int \frac 1 {(x+1)^2 + 16} \,dx = \int \frac{1/16}{\left( \frac{x+1} 4 \right)^2 + 1} \, dx = \int \frac{1/4}{\left( \frac{x+1} 4 \right)^2 + 1} \, \frac{dx} 4 = \frac 1 4 \int \frac{du}{u^2+1} $$ etc.

There is a standard technique in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial without a first-degree term.

Answer 2

Notice that after completing the square,

$$x^2 + 2x + 17 = (x + 1)^2 + 4^2$$

so we can rewrite the integral as a difference:

$$\int \frac{x + 1}{(x + 1)^2 + 4^2} \, dx - \int \frac{1}{(x + 1)^2 + 4^2} \, dx$$

There is a natural substitution in each of these integrals, one leading to a logarithm and the other leading to an arctangent.

Answer 3

You could also do $\int\frac{x+2-2 \mathrm dx}{{x}^2+2x+17}=\int\frac{x+2 \mathrm dx}{{x}^2+2x+17}+\int\frac{-2 \mathrm dx}{{x}^2+2x+17}$

For the first term, let $u=x^2+2x+17$ and $\mathrm du=2x+2 \mathrm dx$.

For the second term, complete the square in the denominator and use the formula for the integral for the form $\frac{1}{a^2+x^2}$. Please comment if you need more elaboration.