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

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Answer 1 · 2018-07-06T06:31:48

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

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#