Question

Integration of 2x+3/x square+2x+1

Answer 1

2 \ln|x+1| - \frac{1}{x+1} + C

Answer 2

The problem asks for the integration of the expression $\frac{2x+3}{x^2+2x+1}$.

First, identify the denominator: $x^2+2x+1$. This is a perfect square trinomial, which can be factored as $(x+1)^2$.

So, the integral becomes: $\int \frac{2x+3}{(x+1)^2} dx$

We can solve this integral using the method of substitution. Let $u = x+1$. Then, differentiating both sides with respect to $x$, we get $du = dx$. Also, from $u = x+1$, we can express $x$ as $x = u-1$.

Now, substitute $u$ and $x$ in terms of $u$ into the integral: $\int \frac{2(u-1)+3}{u^2} du$ Simplify the numerator: $\int \frac{2u-2+3}{u^2} du$ $\int \frac{2u+1}{u^2} du$

Next, split the fraction into two separate terms: $\int \left(\frac{2u}{u^2} + \frac{1}{u^2}\right) du$ $\int \left(\frac{2}{u} + u^{-2}\right) du$

Now, integrate each term separately: The integral of $\frac{2}{u}$ with respect to $u$ is $2 \ln|u|$. The integral of $u^{-2}$ with respect to $u$ is $\frac{u^{-2+1}}{-2+1} = \frac{u^{-1}}{-1} = -\frac{1}{u}$.

Combining these results, the integral in terms of $u$ is: $2 \ln|u| - \frac{1}{u} + C$ where $C$ is the constant of integration.

Finally, substitute back $u = x+1$ to express the result in terms of $x$: $2 \ln|x+1| - \frac{1}{x+1} + C$