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Integration help

xx
(edited 3 years ago)
Reply 1
Original post by NedStark
How do you integrate:

191(x2+6x+11)2-19\int\frac{1}{(x^2+6x+11)^2}

Would appreciate the method first and if possible a solution in spoiler tags!


Haven't got time for a proper look right now, but by completing the square in the denominator and then making a linear translation you can make the denominator look like (u^2 + a^2)^2 for some constant a. Then I think you want to be making some sort of trig or hyperbolic substitution - something like u=atanθu = a\tan \theta. See if that helps :smile:
Reply 2
Original post by NedStark
How do you integrate:

191(x2+6x+11)2-19\int\frac{1}{(x^2+6x+11)^2}

Would appreciate the method first and if possible a solution in spoiler tags!


As davros wrote complete the square

191((x+3)2+2)2dx\displaystyle -19\int \frac{1}{\left ((x+3)^2+2\right )^2} dx

No use the x+3=2tanu x+3=\sqrt{2} tan u substitution

From this x=2tanu3dx=21cos2udu x=\sqrt{2} tan u-3 \rightarrow dx=\sqrt{2} \frac{1}{\cos^2 u} du

Considering that 1tan2u+1=cos2u\frac{1}{tan^2 u+1}=\cos^2 u

191((x+3)2+2)2dx=194cos4u1cos2udu=\displaystyle -19 \int \frac{1}{\left ((x+3)^2+2\right )^2} dx=-\frac{19}{4} \int \cos^4 u \cdot \frac{1}{\cos^2 u} du =

=194cos2udu=194(12u+12cos2u)du\displaystyle =-\frac{-19}{4} \int \cos^2 u du =\frac{-19}{4} \int \left (\frac{1}{2} u +\frac{1}{2}cos 2u\right ) du

Now you can integrate this easily
(edited 8 years ago)

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