Question

Evaluate the integral of 17/(x^(3)-125)

Answer 1

`x^3-125=(x-5)(x^2+5x+25)`


`(17)/((x-5)(x^2+5x+25))=\frac a{x-5}+(bx+c)/(x^2+5x+25)`

`(17)/((x-5)(x^2+5x+25))=(a(x^2+5x+25)+(bx+c)(x-5))/((x-5)(x^2+5x+25))`

`\implies17=(a+b)x^2+(5a-5b+c)x+(25a-5c)`

`\implies\begin{cases}a+b=0\\5a-5b+c=0\\25a-5c=17\end{cases}\implies a=(17)/(75),b=-(17)/(75),c=-(34)/(15)`


`\displaystyle\int(17)/(x^3-125)\,\mathrm dx=(17)/(75)\int(\mathrm dx)/(x-5)-(17)/(75)\int\frac x{x^2+5x+25}\,\mathrm dx-(34)/(15)\int(\mathrm dx)/(x^2+5x+25)`

`\displaystyle=(17)/(75)\int(\mathrm dx)/(x-5)-(17)/(150)\int(2x+5)/(x^2+5x+25)\,\mathrm dx-(17)/(10)\int(\mathrm dx)/(x^2+5x+25)`

`\displaystyle=(17)/(75)\int(\mathrm dx)/(x-5)-(17)/(150)\int(2x+5)/(x^2+5x+25)\,\mathrm dx-(17)/(10)\int\frac{\mathrm dx}{\left(x+\frac52\right)^2+\frac{75}4}`

The first integral is trivial. For the second, replace
`y=x^2+5x+25` so that
`\mathrm dy=(2x+5)\,\mathrm dx`. For the third, take
`x+\frac52=\frac{5\sqrt3}2\tan z` so that
`\mathrm dx=\frac{5\sqrt3}2\sec^2z\,\mathrm dz`.


`\displaystyle=(17)/(75)\ln|x-5|-(17)/(150)\int\frac{\mathrm dy}y-(17)/(10)\int\frac{\frac{5\sqrt3}2\sec^2z}{\left(\frac{5\sqrt3}2\tan z\right)^2+\frac{475}4}\,\mathrm dz`

`\displaystyle=(17)/(75)\ln|x-5|-(17)/(150)\ln|y|-(17)/(25\sqrt3)\int(\sec^2z)/(\tan^2z+1)\,\mathrm dz`

`\displaystyle=(17)/(75)\ln|x-5|-(17)/(150)\ln(x^2+5x+25)-(17)/(25\sqrt3)\int\mathrm dz`

`\displaystyle=(17)/(75)\ln|x-5|-(17)/(150)\ln(x^2+5x+25)-(17)/(25\sqrt3)z+C`

`\displaystyle=(17)/(75)\ln|x-5|-(17)/(150)\ln(x^2+5x+25)-(17)/(25\sqrt3)\arctan(2x+5)/(5\sqrt3)+C`