Evaluate the integral \int \frac{\ln(x^2+2x+2)}{(x+1)^2}\, dx

Question:

Evaluate the integral {eq}\int \frac{\ln(x^2+2x+2)}{(x+1)^2}\, dx {/eq}

Integration By Parts:

Just to recall the formula for the integration by parts method is as follows: {eq}\int \:uv'=uv-\int \:u'v {/eq} This formula is used in the composite functions that is in the expression of the integral. Now once this formula is applied the integral will be simplified and then to evaluate we can use other methods if required.

Answer and Explanation:


The given integral is :

{eq}\int \frac{\ln(x^2+2x+2)}{(x+1)^2} dx\\ {/eq}

Integrating by parts, let:

{eq}u=\ln \left(x^2+2x+2\right),\:v'=\frac{1}{\left(x+1\right)^2}\\ \Rightarrow -\frac{\ln \left(x^2+2x+2\right)}{x+1}-\int \:-\frac{2}{x^2+2x+2}dx\\ =-\frac{\ln \left(x^2+2x+2\right)}{x+1}+ 2\cdot \int \:\frac{1}{\left(x+1\right)^2+1}dx\\ =-\frac{\ln \left(x^2+2x+2\right)}{x+1}+2\arctan \left(x+1\right)+C {/eq}


Learn more about this topic:

Using Integration By Parts

from Math 104: Calculus

Chapter 11 / Lesson 7
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