integral(x/(x^2+4x+13),x,0,1) I broke the bottom up to (x+2)^2 + 9, but now I"m stuck. any help would be very appreciated. Thanks, s3n4te edit: i think i should now use inverse trig substitution.
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Hi $\displaystyle \frac{x}{x^2+4x+13} = \frac{\frac{1}{2}\2x+4)}{x^2+4x+13} - \frac{2}{(x+2)^2+9}$ The first part is integrable in $\displaystyle \frac{1}{2}\:ln|x^2+4x+13|$ The second part is integrable in Arctan after setting x+2=3t
How do you know how to split that up like that?
Originally Posted by s3n4te How do you know how to split that up like that? By experience You make $\displaystyle \frac{u'}{u}$ appear eventually using a multiplicative constant - here $\displaystyle \frac{1}{2}$ The other part is then integrable with Arctan (this is true because the denominator has no real roots)
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