Why these integrals have different results ?: $A=\int_{-5}^0 \frac{1}{\left(x^2+2^2\right)^{3/2}} \,dx$ $\{-x,0\}$ and $B=\int_{-5}^0 \frac{-x}{\left(x^2+2^2\right)^{3/2}} \, dx$. Vector $\overset{\to }{u}=\{-x,0\}=-x\overset{\to }{i}+\overset{\to }{0j}$ . In B, I introduced the component ${-x}$ of $\overset{\to }{u}$ into de Integral to then multiply by $\overset{\to }{i}$ , Is it mathematicaly right? or, How I must solve A?
Evaluating in Mathematica these results are: $A=\left\{-\frac{5 x}{4 \sqrt{29}},0\right\}$ and $B=(\frac{1}{2}-\frac{1}{\sqrt{29}})$. *(it is $B=(\frac{1}{2}-\frac{1}{\sqrt{29}})\overset{\to }{i}$)
EDIT: in this example, it shows they are solving as I solved B: image
You can’t take the $x$ inside the integral when multiplying by a vector, as the integral is a scalar and must be evaluated independently. So $A\ne B$. Consider $A$ as an indefinite integral first. Substitute $x=2\tan y \implies dx = 2\sec^2 y dy $ in $A$ to get $$\int \frac{2\sec^2y}{8\sec^3 y} dy = \frac 14 \int \cos y dy = \frac 14\sin y + c = \frac{x/2}{4\sqrt{1+(x/2)^2}} + c = f(x)$$
Then, just evaluate $f(x)\big|_{-5}^0=A$ and then your answer would be $\big(- Ax,0\big ).$