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# Integration problems watch

1. Im having trouble integrating these 2:

∫x2 /x2 +x-20.dx
and
∫(x2 +1/3)/(x3+x-1)dx

Thanks
2. (Original post by wibble...)
1.) ∫x2 /x2 +x-20.dx
2.)∫(x2 +1/3)/(x3+x-1)dx
1.) ∫x²/(x² + x - 20) dx
= ∫x²/[(x + 5)(x - 4)] dx
= ∫A + B/(x + 5) + C/(x - 4) dx
= ∫{A(x + 5)(x - 4) + B(x - 4) + C(x + 5)}/[(x + 5)(x - 4)] dx

Equating numerators:
=> A(x + 5)(x - 4) + B(x - 4) + C(x + 5) = x²
Let x = 4: => 9C = 16 => C = 16/9
Let x = -5: => -9B = 25 => B = -25/9
Let x = 0: -20A - 4B + 5C = 0
=> -20A + 4(25/9) + 5(16/9) = 0
=> -20A + 100/9 + 80/9 = 0
=> -20A = - 180/9 = 20
=> A = -1

:. ∫x²/(x² + x - 20) dx
= ∫-1 - 25/[9(x + 5)] + 16/[9(x - 4)] dx
= (-1/9) ∫9 + 25/(x + 5) - 16/(x - 4) dx
= (-1/9){9x + 25ln|x + 5| - 16ln|x - 4|} + k

2.) ∫(x² + 1/3)/(x³ + x - 1) dx
= (1/3) ∫ (3x² + 1)/(x³ + x - 1) dx
= (1/3) ln|x³ + x - 1| + k
3. (Original post by wibble...)
Im having trouble integrating these 2:

∫x2 /(x2 +x-20).dx
you need to rewrite it in partial fractions before you integrate.

=>∫5/{9(x + 5)} +∫4/{(x - 4)} which will no doubt require ln....

∫(x2 +1/3)/(x3+x-1)dx is easier...the top is 1/3 of the differential of the bottom so you can write the answer straight away as

1/3ln|x3+x-1| + c
4. your answer for 1) isn't right you can't use partial fractions if the numerator is of the same order as the denominator
5. Complete the square on the bottom line, giving

(x + ½)² - 20.25.

Then let u = x + 1/2, so you get

(u-0.5)^2/u^2 - 20.25 du

= ∫u^2 - u + 0.25/u^2 - 20.25 du

= (∫u^2/u^2 - 20.25 du) - (∫u/u^2 - 20.25 du ) +0.25( ∫1/u^2 - 20.25 du)

Problem should be soluble from there using standard techniques

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