How do you integrate #int 1 / (xsqrt(4x^2 +9))dx# using trigonometric substitution?

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Answer 1 · 2018-03-21T12:38:25

#int dx/(xsqrt(4x^2+9)) = 1/6 ln abs (3-sqrt(4x^2+9)) -1/6 ln (3+sqrt(4x^2+9))+ C#

Substitute:

#u = sqrt(4x^2+9)#

#du = (4xdx)/sqrt(4x^2+9)#

Then:

#int dx/(xsqrt(4x^2+9)) = 1/4 int 1/x^2(4xdx)/sqrt(4x^2+9)#

and as:

#x^2 = (u^2-9)/4#

then:

#int dx/(xsqrt(4x^2+9)) = 1/4 int 4/(u^2-9) du = int (du)/((u-3)(u+3))#

Using now partial fractions decomposition:

#1/((u-3)(u+3)) = A/(u-3)+B/(u+3)#

#1 = A(u+3)+B(u-3)#

#1 = (A+B)u+3(A-B)#

#{(A+B=0),(A-B=1/3):}#

#{(A=1/6),(B=-1/6):}#

So:

#int dx/(xsqrt(4x^2+9)) = 1/6 int (du)/(u-3) -1/6 int (du)/(u+3)#

#int dx/(xsqrt(4x^2+9)) = 1/6 ln abs(u-3) -1/6 ln abs (u+3) +C#

and undoing the substitution:

#int dx/(xsqrt(4x^2+9)) = 1/6 ln abs (3-sqrt(4x^2+9)) -1/6 ln (3+sqrt(4x^2+9))+ C#