3 years ago

I need help. ASAP! Divide. (3b^3 – 10b^2 + 4) ÷ (3b – 1)

Mathematics

1 answer:

viktelen [127]3 years ago

3 0

Step-by-step explanation:

try surfing how to divide a function by the long division method

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Find the value of c and YZ if Y is between X and Z.<br><br><br><br> XY=5.5, YZ=2c, XZ=8.9

andrey2020 [161]

Answer:

XZ = XY+ YZ

8.9=5.5+2c

2c= 8.9-5.5

2c=3.4

c=1.7

YZ = 2c=3.4

Step-by-step explanation:

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3 years ago

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

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

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$