Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We want$$\int \dfrac{2x}{4x^2+1}$$

I only know that $\ln(4x^2 + 1)$ would have to be in the mix, but what am I supposed to do with the $2x$ in the numerator?

share|cite|improve this question
up vote 4 down vote accepted

Hint: make the substitution $u = 4x^{2} +1$. Then $du = 8xdx$, and the integral becomes:

$$\frac{1}{4} \int \frac{8xdx}{4x^{2}+1} = \frac{1}{4} \int \frac{du}{u}$$

share|cite|improve this answer

Try differentiating $\ln(4x^2+1)$ and see what you get (don't forget the chain rule!). Then see how this compares to the integrand.

share|cite|improve this answer

Again, as in your past question, there's a general case here: if $\,f\,$ is derivable then

$$\int\frac{f'(x)}{f(x)}dx=\log|f(x)|+K$$

Here, we have

$$\frac{2x}{4x^2+1}=\frac14\frac{(4x^2+1)'}{4x^2+1}\ldots$$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.