Integration help for a beginner

Question

This might be a silly question but I am a beginner in calculus so I really do not understand. When solving the problem $\int\frac{5+x}{\sqrt{16−(x+4)^2}}dx$, why we can’t divide it into the two integrals $\int\frac{5}{\sqrt{16−(x+4)^2}}dx$ and $\int\frac{x}{\sqrt{16−(x+4)^2}}dx$ and just add them? I am really struggling with this.

Answer 1

Integrals of this sort are most of the times solved with u-sub. If we substitute : $u=x+4$ we get $du=dx$. The integral becomes :

$$\int\frac{5+x}{\sqrt {16-u^2}}du$$

We note that $u=x+4\Rightarrow x=u-4$ which makes things much easier as we get :

$$\int\frac{u+1}{\sqrt {16-u^2}}du=\int\frac{u}{\sqrt {16-u^2}}du+\int\frac{1}{\sqrt {16-u^2}}du$$

For the first one we again use u-sub with : $v=16-u^2\Rightarrow \frac{dv}{du}=-2u$ and the integral becomes : $$\int\frac{u}{\sqrt {16-u^2}}du=\int\frac{u}{\sqrt {v}}\frac{1}{-2u}dv=-\int\frac{1}{2\sqrt {v}}dv=-\sqrt{v}+C_2$$

The second one is a standard one : We substitute $p=\frac{u}{4}\Rightarrow \frac{dp}{du}=\frac{1}{4}$ and we get : $$\int\frac{1}{\sqrt {16-u^2}}du=\int\frac{4}{\sqrt {16-16p^2}}dp=\int\frac{1}{\sqrt {1-p^2}}dp=\arcsin(p)+C_3$$

Now we have our final result : $$\int\frac{5+x}{\sqrt {16-(x+4)^2}}du=-\sqrt{v}+\arcsin(p)$$$$=-\sqrt{16-(x+4)^2}+\arcsin{(\frac{x+4}{4})}+C$$

Integrals are linear operators : so in general the following holds for $f,g:\mathbb{R}\rightarrow \mathbb{R}$ who are integrable : $$\int(af(x)+bg(x))dx=a\int f(x)dx+b\int g(x)dx$$

Answer 2

Before splitting the integrand, let’s try the other substitution $$ x+4=4 \sin \theta \text {, then } d x=4 \cos \theta d \theta $$$$ \begin{aligned} \int \frac{5+x}{\sqrt{16-(x+4)}} & =\int x=\int \frac{1+4 \sin \theta}{\sqrt{16-16 \sin ^2 \theta}} 4 \cos \theta d \theta \\ & =\int(1+4 \sin \theta) d \theta \\ & =\theta-4 \cos \theta+C \\ & =\arcsin \left(\frac{x+4}{4}\right)-\sqrt{16-(x+4)^2}+C \end{aligned} $$