# Thread: Another Integration Problem - Understanding Numerator

1. ## Another Integration Problem - Understanding Numerator

$\int \dfrac{x + 3}{x^{2} + 2x + 2}dx$

$\dfrac{1}{2} \int \dfrac{(x + 1) 2dx}{(x + 1)^{2} + 1} + \int \dfrac{2dx}{(x + 1)^{2} + 1}$ - What's going on with the numerator? I understand the denominator (completing the square).

For instance, how can $dx$ be fixed (u substitution)? Given: $u = x^{2} + 2x + 2$ and $du = 2x + 2$. Note: $du$ and $dx$ have to be fixed before splitting the fraction into two fractions.

$\dfrac{1}{2} \ln(x^{2} + 2x + 2) + 2 \arctan(x + 1) + C$ - Final answer

2. ## Re: Another Integration Problem - Understanding Numerator

Jason
It seems to me that you lack basic knowledge about integration.
I suggest you download asap a calculus book and start to revise the chapter on integration.
you may go here : Pauls Online Math Notes

Now some hints...if u=x^2+2x+2 then du =(2x+2)dx therefore dx=(du)/(2x+2). du is the differential of the function u and as such it is equal to the derivative of u times the diferrential of x which is dx...
There is nothing wrong with the numerator....the first integral can be manipulated to become ( the numerator) d[(x+1)^2+1] and as such it will give the ln(x^2+2x+2) after integration. the second one is related to the inverse trigonometric functions...do you have any idea of what I am talking about? please revise the related topic.
Anyway the answer you post is correct.

3. ## Re: Another Integration Problem - Understanding Numerator

\begin{align*}&\int \frac{x + 3}{x^{2} + 2x + 2}dx\\ =& \int \left( \frac{2}{x^2 + 2x + 2} + \frac{x+1}{x^{2} + 2x + 2}\right)\,\,dx...\text{[by algebra]} \\ =& \int \frac{2}{x^2 + 2x + 2}\,\, dx + \int \frac{x+1}{x^{2} + 2x + 2}\,\,dx....\text{[by property of integration]} \\=& 2 \int \frac{1}{(x + 1)^2 + 1^2}\,\,dx + \frac{1}{2} \int \frac{2(x+1)}{x^{2} + 2x + 2}\,\,dx\\ \\& \text{Let } u = x+1 \,\,\therefore du = dx\\& \text{And Let } m = x^{2} + 2x + 2\,\, \therefore dm = 2(x+1)\,\,dx\\\\=& 2\int \frac{1}{u^2 + 1^2}\,\,du + \frac{1}{2}\int \frac{1}{m}\,\,dm....\text{[by plugging in } m, dm, u, du\text{ ]}\\=& 2\tan^{-1}(u) + \frac{1}{2}\ln|m| + C\\=& 2\tan^{-1}(x+1) + \frac{1}{2}\ln|x^2 + 2x + 2| + C....\text{[by substituting } u \text{ and } m\text{ ]}\end{align*}

Hope it helps.