<s>[INST] Solving an integral I am having trouble with this integral:
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx.$$
One obvious thing would be to complete the square: $x^2+2x+25=(x+1)^2+24$. But then, I don't know which substitution to use. Can anyone help? Thank you.
 [/INST] You will have, by substitution of $y=\frac{x+1}{\sqrt{24}}$,
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy$$
that becomes
$$\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy=\sqrt{24}\int_\frac{1}{\sqrt{24}}^1\frac{y}{\sqrt{y^2+1}}dy-\int_\frac{1}{\sqrt{24}}^1\frac{1}{\sqrt{y^2+1}}dy.$$
This gives finally
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\sqrt{48} - \sqrt{26} + {\rm arcsinh}\left(\frac{1}{\sqrt{24}}\right) -  {\rm arcsinh}(1).$$
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