Value of ₀¹ 1/(x+2)(x+3) \, dx is

Correct Answer is : log⁡(98)≤ft(9/8)log(89)

Correct Answer is : log⁡(98)≤ft(9/8)log(89)

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Integral Calculus

Question : 22 of 73

Marks: +1, -0

Value of

\int\limits_{0}^{1} \frac{1}{(x+2)(x+3)} \, dx

Solution:

Concept:

Use partial fractions to integrate the rational function.

Explanation:

Write the integrand as partial fractions:

\frac{1}{(x+2)(x+3)} = \frac{1}{x+2} - \frac{1}{x+3}

Integrate term by term:

\int \frac{1}{x+2} \, dx = \ln|x+2|

and

\int \frac{1}{x+3} \, dx = \ln|x+3|

So the definite integral is:

\left[ \ln|x+2| - \ln|x+3| \right]_{0}^{1} = \left[ \ln\left(\frac{x+2}{x+3}\right) \right]_{0}^{1}

Evaluate at limits:

x=1

\ln\left(\frac{3}{4}\right)

x=0

\ln\left(\frac{2}{3}\right)

Subtract:

\ln\left(\frac{3}{4}\right) - \ln\left(\frac{2}{3}\right) = \ln\left( \frac{3/4}{2/3} \right) = \ln\left( \frac{3}{4} \cdot \frac{3}{2} \right) = \ln\left(\frac{9}{8}\right)

Answer:

\log\left(\frac{9}{8}\right)

(Option B).

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