Solving an integral with u substitution brings an integral into elementary form through a change of variable technique. In this example we consider a problem where it may not be obvious how to apply the method. Consider:

int{}{}{{dx} /{4 x^2 ~+~8 x ~+~13}}

We’ll see in a moment that this integral evaluates to an inverse tangent. We’re going to use two steps to convert this into the following elementary form:

int{}{}{{dx} /{ x^2~ +~1}}~=~tan^-1x~+~C

The first step is to bring the denominator into a more appropriate form. We want something like:

a x^2~ +~ b

where a and b are constants. When you see a problem with a quadratic function in the denominator, try completing the square. In this case we can do the following:

4x^2~+~8x~+~13~=~4(x^2~+~2x)~+~13

~=~4[x^2~+~2x~+(1)^2]~+~13~-~4(1)^2

~=~4(x+1)^2~+~9

Now we can write the integral as follows:

int{}{}{{dx} /{4 (x~+~1)^2 ~+~9}}

Now we can to the integral by u substitution. To get it in the form we want, where we can get the form for the inverse tangent integral, we will have to pick u such that we can pull out all the constants in the denominator. This can be done if we take:

9u^2~=~4(x+1)^2

We do this because we could transform the integral as follows:

int{}{}{{dx} /{4 (x~+~1)^2 ~+~9}}~doubleright~int{}{}{{du} /{9u^2 ~+~9}}

modulo a proportionality factor. Taking the square root of our chosen u substitution, we get:

3u~=~2(x+1)

That is:

3du~=~2dx,~doubleright~dx~=~{3/2}du

Now we can get the result we seek by using this substitution in the original integral. This gives:

int{}{}{{dx} /{4 (x~+~1)^2 ~+~9}}~=~{3/2}int{}{}{{du} /{9u^2 ~+~9}}

~=~{3/2}{1/9}int{}{}{{du} /{u^2 ~+~1}}

~=~{1/6}int{}{}{{du} /{u^2 ~+~1}}~=~{1/6}tan^-1u~+~C

Now replace u to get the final result:

int{}{}{{dx} /{4 x^2 ~+~8 x ~+~13}}~=~{1/6}tan^-1{2/3}(x+1)~+~C