# Thread: Help integrating this simple function.

1. ## Help integrating this simple function.

I have this function: f(x) = (4x-3)/(x² + 1) (upper limit is 3/4, lower limit is 0)

i have trouble integrating it; if it were simply f(x) = 4x/(x² +1) then that would be easy but the -3 causes problems...

in my solutions it has a really strange explanation of "argtgx" which i have never seen before, help very very appreciated please!

2. Rewrite it as $\displaystyle \displaystyle \int{\frac{4x}{x^2 + 1} - \frac{3}{x^2 + 1}\,dx}$.

The first term is integrated using a $\displaystyle \displaystyle u$ substitution, while the second term is integrated using the substitution $\displaystyle \displaystyle x = \tan{\theta}$.

3. Originally Posted by Prove It
Rewrite it as $\displaystyle \displaystyle \int{\frac{4x}{x^2 + 1} - \frac{3}{x^2 + 1}\,dx}$.

The first term is integrated using a $\displaystyle \displaystyle u$ substitution, while the second term is integrated using the substitution $\displaystyle \displaystyle x = \tan{\theta}$.
I get the first term, but honestly, having done calculus for years I've NEVER seen an integration/differentiation using tan, how is that done? i only use cos and sin...

4. Make the substitution $\displaystyle \displaystyle x = \tan{\theta}$. What is $\displaystyle \displaystyle \frac{dx}{d\theta}$?

5. note ...

$\displaystyle \dfrac{d}{dx} \arctan{x} = \dfrac{1}{x^2+1}$