integral [(x+3)/((x+5)sqrt(x+4)] dx?

Answer 1

-2*(2*arctan(sqrt(x+ 4))-sqrt( x + 4))

Answer 2

New and improved:

(X+3)/[(x+5)* sqrt(x+4)], let y = x+4 then dy = dx

(y - 1)/[(y + 1)* sqrt(y)] =

y/[(y + 1)* sqrt(y)] - 1/[(y+1)* sqrt(y)]
work with this 1st work with this second

1st formula: which is

sqrt(y)/(y+1) , now let y = Z^2 then dy = 2z dz

2z^2/ (z^2 + 1), when you do the division we get

2 - 2/(z^2 + 1), then we integrate
intergral 2 dz - integral 2/(z^2 + 1) dz

intergral 2 dz -Integral 2/(z^2 + 1)dz = Integral 2 dz - 2Arctan(z), but z = sqrt(y) = sqrt(X+4)
there for the 1st formula becomes

2 sqrt(x+4) - 2Arctan(sqrt(x+4)) + C1

Now look at the second formula, ignoring for now the -1 coeff

1/[(y+1)*sqrt(y)] dy, again let y = z^2 then dy = 2z dz

2z/[(Z^2 + 1)*z] dz= 2/(z^2 + 1] dz

integrate 2/(z^2 + 1) dz = 2*Arctan(z) = 2Arctan(sqrt(y)) =
2Arctan(sqrt(x+4)) + c2

Now remember the -1 coeficient and add both formulas

2sqrt(x+4) - 4Arctan(sqrt(x+4)) + C


It is not impossible that I could have made a mistake, I am abit too lazy to check, a derivative should give the original problem

Answer 3

3

Answer 4

⌠ - Integral sign.

⌠(x+3)/((x+5)²(x+4)) dx (Partical fractions).

⌠a/(x+5) + ⌠b/(x+5)² + ⌠c/(x+4) dx

⌠1/(x+5) + ⌠2/(x+5)² + ⌠-1/(x+4) dx

⌠1/(x+5) + ⌠2/(x+5)² - ⌠1/(x+4) dx

⌠1/(x+5) + 2⌠(x+5)^-² - ⌠1/(x+4) dx

ln(x+5) + [2(x+5)^-1]/-1 - ln(x+4) + c

ln(x+5) - 2/(x+5) - ln(x+4) + c

Answer 5

using the program "Maple 10" ...

The answer is : 2*sqrt(x+4)-4*arctan(sqrt(x+4))

Don't know how to solve it by myself. Maybe if you say something like....sqrt(x+4) = sin(x) . But, really didn't try it. :-)

Well just use the answer that Maple gives.

Answer 6

Dont know too complicated

Answer 7

what are the limits?

the old school method is like this

[dinominator*derivative of num -- num*deriv. of din.]
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[dinominator]^2