# Help finding the area under a curve

I was given an excercise in my calculus class that i don't really understand, the problem says : Find the area limited by the curves $$y = \frac{x+4}{x^2+1} ,\space x = -2 ,\space x = 3,\space y = 0$$

I don't really know what approach to follow here, my guess would be to solve it using riemann sums or maybe definite integrals and using $x = -2$ and $x = 3$ as the interval but i'm totally lost.

• It is almost always best to start these types of questions by sketching the curves, finding the intersection points, and figuring out which area you want to find. Then you can move on to deciding how to integrate. – Zubin Mukerjee Jan 15 '15 at 20:19

Yes, this is a definite integral problem. You can solve it by evaluating

$$\int_{-2}^3{x+4\over x^2+1}\,dx.$$

Hint: Split it into

$$\int_{-2}^3\frac x{x^2+1}\,dx+4\int_{-2}^3\frac1{x^2+1}\,dx,$$

and recall that the derivative of $\tan^{-1}x$ is $\frac1{x^2+1}$.

• you are missing a $4$ in the second integrand. – abel Jan 15 '15 at 20:23
• Awesome, thank you. I solved it this way and got the desired result – HardCodeStuds Jan 15 '15 at 20:28

It is meant that you should calculate the integral $$\int_{-2}^3 \frac{x+4}{x^2+1}\,dx$$ Also, see this picture where the curves are drawn.

Let $u = x^2+1$. The derivative of $u$ with respect to $x$ is $2x$. If $A$ is the area we want, then

\begin{align} A= \int_{-2}^{3} \frac{x+4}{x^2+1}\,\mathrm{d}x&=\frac{1}{2}\left(\int_{-2}^{3} \frac{2x}{x^2+1} \,\mathrm{d}x\right)+4\left(\int_{-2}^{3} \frac{1}{x^2+1}\,\mathrm{d}x\right) \\\\ &=\frac{1}{2}\left(\int_{5}^{10} \frac{\mathrm{d}u}{u}\right)+4\left(\int_{-2}^{3} \frac{1}{x^2+1}\,\mathrm{d}x\right) \end{align}

Remember, as Tim Raczkowski tells us, that $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\arctan x\right) = \frac{1}{x^2+1}$$

Also:

$$\frac{\mathrm{d}}{\mathrm{d}x} \left(\ln \left|x\right|\right) = \frac{1}{x}$$

This gives

\begin{align} A&=\frac{\ln 10 - \ln 5}{2} +4 \left(\arctan 3 - \arctan \left(-2\right)\right)\\\\ &\approx\boxed{9.7714} \end{align}