Evaluate 2x+1/x(x+1) \, dx

Correct Answer is : log⁡(x2+x)+c(x² + x) + clog(x2+x)+c

Correct Answer is : log⁡(x2+x)+c(x² + x) + clog(x2+x)+c

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Integral Calculus

Question : 67 of 73

Marks: +1, -0

Evaluate

\int \frac{2x+1}{x(x+1)} \, dx

Solution:

Concept:

Decompose the integrand into partial fractions to integrate term by term.

Explanation:

Write

\frac{2x+1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}

Multiplying both sides by

x(x+1)

gives

A(x+1) + Bx = 2x+1

Set

x=0

A = 1

Set

x=-1

-B = -1

, so

B=1

Thus the integrand is

\frac{1}{x} + \frac{1}{x+1}

Integrate:

\int \frac{1}{x}\,dx + \int \frac{1}{x+1}\,dx = \ln|x| + \ln|x+1| + c = \ln|x(x+1)| + c = \ln|x^2+x| + c

Answer:

\log(x^2+x) + c

(option B).

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