integrations-by fraction decomp. and trig subst.

[hgaon001]

The first one says

integrand of (2x^2+13)/[(x^2+4)(x-1)]

I did integration by fraction decomposition.. I got B=3 and A=-1.. But when i plug in to the original of (2x^2+13)=B/(x-1)+A/(x^2+4)im not gettin back to the original answer..

and the other one reads

Integrand from 4 to 8 of [sqrt(x^2-16)]/x^2

i got as far as a=4 and u=x=4sintheta dx=4costhetadtheta

after pluggin in i got stuck at the intregrand of [(4costheta)(4costheta)dtheta]/16sintheta

 

[DrMike]

The first one says

integrand of (2x^2+13)/[(x^2+4)(x-1)]

I did integration by fraction decomposition.. I got B=3 and A=-1.. But when i plug in to the original of (2x^2+13)=B/(x-1)+A/(x^2+4)im not gettin back to the original answer..

When you have a quadratic term on the bottom, you need a linear term on top. The correct decomposition has the form

\(\displaystyle \frac {2x^2+13}{(x^2+4)(x-1)} = \frac {Ax+B}{x^2+4} + \frac{C}{x-1}\)

Integrand from 4 to 8 of [sqrt(x^2-16)]/x^2

i got as far as a=4 and u=x=4sintheta dx=4costhetadtheta

after pluggin in i got stuck at the intregrand of [(4costheta)(4costheta)dtheta]/16sintheta

I think there's a couple of errors here... firstly,

\(\displaystyle \sqrt{16 \sin^2(\theta)-16}=4\sqrt{-\cos^2(\theta)}\)

so you have an imaginary problem - that is, a very real problem that imaginary numbers are creeping in.

secondly, doesn't the x[sup:1ix7k1fo]2[/sup:1ix7k1fo] become 16sin[sup:1ix7k1fo]2[/sup:1ix7k1fo](theta), instead of 16sin(theta) ?

You'll have better luck with either x=4 sec(theta) or x=4 cosh(t)

 

[fasteddie65]

Just to clarify the previous suggestions:

For quadratic denominators THAT ARE PRIME (i.e. not factorable), the numerator must be linear.

If a quadratic denominator is factorable, then break it down into its linear factors, and use the usual constant numerator as you did in the problem you showed.

In the second problem, the sec substitution is probably more familiar than the cosh one, so I always use a sec ø for an expression like x^2 - a^2.

 

[BigGlenntheHeavy]

\(\displaystyle \int_{4}^{8}\frac{\sqrt(x^{2}-16)}{x^{2}}dx \ Let \ x \ = \ 4sec\theta, \ then \ x^{2} \ = \ 16sec^{2}\theta \ and \ dx \ = \ 4sec\theta \ tan\thetad\theta d\theta.\)

\(\displaystyle This \ reduces \ to \ \int_{0}^{\frac{\pi}{3}}tan\theta sin\theta d\theta.\)

\(\displaystyle Now, \ \int_{0}^{\frac{\pi}{3}}tan\theta sin\theta d\theta \ = \int_{0}^{\frac{\pi}{3}}(-cos\theta+sec\theta)d\theta\)

\(\displaystyle Which \ equals \ [-sin\theta+ln|sec\theta+tan\theta|]_{0}^{\frac{\pi}{3}} \ = \ ln(2+\sqrt3)-\frac{\sqrt3}{2}.\)

Note: x^2-16 ? 0, |x| ? 4, hence x ? 4 (which it is) or x ? -4 (not in the domain).

Another note as I feel like using LaTex.

\(\displaystyle Identity: \ tan\theta \ sin\theta \ = \ \frac{sin^{2}\theta}{cos\theta} \ = \ \frac{1-cos^{2}\theta}{cos\theta} \ = \ \frac{1}{cos\thets} \ - \ cos\theta \ = \ sec\theta \ - \ cos\theta.\)