Simple İntegral question Check

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Homework Statement


##\int \frac{x}{\left(C-x\right)^2}dx=?##
Here ##C## is a constant

Homework Equations

The Attempt at a Solution


##c-x=u## then ##dx=-du##
then it becomes,
##\int \frac{u-c}{\left(u\right)^2}du##
##\int \frac{u}{u^2}-\frac{c}{u^2}du##
From there I found
##ln\left(c-x\right)+\frac{c}{c-x}##
but symbolab says its,
##ln\left(c-x\right)+\frac{x}{c-x}##
I cannot see how ??
 
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It's probably quicker to just differentiate your result than to check it with software from which you learn nothing.
 
alan2 said:
It's probably quicker to just differentiate your result than to check it with software from which you learn nothing.
Make sense never thought.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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