Evaluate: 1/x(x+1) \, dx

Correct Answer is : log⁡x−log⁡(x+1)+c x - (x+1) + clogx−log(x+1)+c

Correct Answer is : log⁡x−log⁡(x+1)+c x - (x+1) + clogx−log(x+1)+c

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Integral Calculus

Question : 73 of 73

Marks: +1, -0

Evaluate:

\int \frac{1}{x(x+1)} \, dx

Solution:

Concept:

Decompose the integrand into partial fractions.

Explanation:

Write

\frac{1}{x(x+1)}

\frac{A}{x} + \frac{B}{x+1}

Multiply both sides by

x(x+1)

1 = A(x+1) + Bx

Set

x = 0

gives

1 = A(1) \Rightarrow A = 1

Set

x = -1

gives

1 = B(-1) \Rightarrow B = -1

Thus

\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}

Integrate:

\int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \log|x| - \log|x+1| + c

Answer:

\log x - \log(x+1) + c

(with absolute values implied).

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