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3 years ago

PLEASE HELP THE CORRECT ANSWER WILL BE MARKED BRAINLIST!

Mathematics

2 answers:

Lena [83]3 years ago

4 0

Well first you'd need to turn the fractions to improper, for instance you'd need to turn 1 1/3 from a mixed number to an improper fraction you do this by, keeping your denominator and then multiplying 3x1 and adding the 1, after that your final fraction should be 4/3 you also do this for your other fractions, let's remember that we still need to make 17 5/6 improper and 10 2/5 improper. so 17 5/6 will turn into 107/6 we are basically just doing 17x6 plus 5, now 10 2/5 will turn into 52/5 then we will have to simplify our fractions and subtract

Anna11 [10]3 years ago

3 0

Answer:

hi iam coming from india please support me and follow me please mark me brainlist

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vampirchik [111]

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2 years ago

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rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
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3 years ago

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swat32

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