# How to do the integration by parts of 1/(x(x+1))dx?

### 4 Answers

- MyRankLv 61 month ago
1/(x(x+1))

1/x(x+1) = A/x + B/(x+1)

1 = A(x+1) + Bx

Put x = 0

1 = A(0 + 1) + B(0)

1 = A

Put x = -1

1 = A(-1 + 1) + B(-1)

1 = -B

B=-1

1/x(x+1) = 1/x - 1/(x+1)

Integration on both sides

∫1/x(x+1). dx =∫1/x.dx - ∫1/(x+1).dx

= log|x| - log|x+1| + logc

= log|(x/(x+1)|c

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- az_lenderLv 71 month ago
Yet another way, besides partial fractions, would be to rewrite the integrand as

dx/[(x+1/2)^2 - (1/4)], then set

x + 1/2 = (1/2)sec(u),

so dx = (1/2)sec(u)tan(u) du, and

(x+1/2)^2 - (1/4) = (1/4)[sec^2(u) - 1] = (1/4)tan^2(u).

Now the integrand becomes

(1/2)sec(u)tan(u) du/[(1/4)tan^2(u)]

= 2 sec(u) du/tan(u) = 2 csc(u) du,

and after integration you have

-2 ln|csc(u) + cot(u)| + C.

To get back to x, draw a triangle where u is the angle,

x+1/2 is the hypotenuse, and (1/2) is the "adjacent" side.

Then the opposite side is sqrt(x^2 + x), so the above logarithmic expression can be rewritten in "x".

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- 1 month ago
You don't integrate this by parts. You use partial fraction decomposition

1 / (x * (x + 1)) = a/x + b/(x + 1)

a/x + b/(x + 1) =>

a * (x + 1) / (x * (x + 1)) + b * x / (x * (x + 1)) =>

(a * (x + 1) + b * x) / (x * (x + 1)) =>

(ax + bx + a) / (x * (x + 1)) =>

((a + b) * x + a) / (x * (x + 1))

0x + 1 = (a + b) * x + a

a + b = 0

a = 1

1 + b = 0

b = -1

Now you have:

1 / x - 1 / (x + 1)

dx / x - dx / (x + 1)

Integrate

ln|x| - ln|x + 1| + C =>

ln|x / (x + 1)| + C

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- ted sLv 71 month ago
you don't !...partial fraction decomposition...1/x - 1 / (x + 1 )

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