Question

Integration by completing the square. The integral of y equals to the fraction of 1 over 5x squared + 25x-10?

Answer 1

`int\ dx/(5x^2+25x-10)=sqrt(33)/165ln((2x+5-sqrt(33))/(2x+5+sqrt(33)))+C`

Explanation:

I assume that you're asking about `int\ dx/(5x^2+25x-10)`.

First, factor out the `5` in the denominator to prepapre for completing the square:
`int\ dx/(5(x^2+5x-2)`

Now, complete the square:
`1/5int\ dx/((x+5/2)^2-33/4)`

Now, substitute `u=x+5/2` and `dx=du`:
`1/5int\ (du)/(u^2-33/4)`

Now, factorize the denominator to prepare for partial fractions:
`1/5int\ (du)/((u+sqrt(33)/2)(u-sqrt(33)/2))`

Applying partial fractions, we have
`1/5int\ (1/(sqrt(33)(u-sqrt(33)/2))-1/(sqrt(33)(u+sqrt(33)/2)))du`

Which leads to, after some simplifying:
`sqrt(33)/165(ln(u-sqrt(33)/2)-ln(u+sqrt(33)/2))+C`

Or, using the rules of logarithms:
`sqrt(33)/165ln((u-sqrt(33)/2)/(u+sqrt(33)/2))+C`

Finally, substitute `u=x+5/2` back:
`sqrt(33)/165ln((2x+5-sqrt(33))/(2x+5+sqrt(33)))+C`