How do you find the integral of #int(2x)/(x^2+6x+13) dx#

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Answer 1 · 2015-02-16T15:33:27

The answer is: #ln(x^2+6x+13)-3arctan(1/2(x+3))+c#

Follow my passages:

#int(2x)/(x^2+6x+13)dx=int(2x+6-6)/(x^2+6x+13)dx=#

#=int(2x+6)/(x^2+6x+13)dx-6intdx/(x^2+6x+13)=(1)#

#x^2+6x+13=x^2+6x+9+4=(x+3)^2+4=#

#=4[1/4(x+3)^2+1]=4[(1/2(x+3))^2+1]#.

So:

#(1)=ln(x^2+6x+13)-6intdx/(4[(1/2(x+3))^2+1])=#

#=ln(x^2+6x+13)-6/4intdx/((1/2(x+3))^2+1)=#

#=ln(x^2+6x+13)-3/2*2int(1/2)/((1/2(x+3))^2+1)dx=#

#=ln(x^2+6x+13)-3arctan(1/2(x+3))+c#