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Integral Calculus

Question : 10 of 73

Marks: +1, -0

Value of

\int\limits\; \frac{3x+4}{6x+7} dx

Solution:

Concept:

Use polynomial division to simplify the integrand into a constant plus a proper rational function.

Explanation:

Let

3x+4 = A(6x+7) + B

Comparing coefficients:

6A = 3 \implies A = \frac{1}{2}

;

7A + B = 4 \implies \frac{7}{2} + B = 4 \implies B = \frac{1}{2}

Thus

\frac{3x+4}{6x+7} = \frac{1}{2} + \frac{1}{2(6x+7)}

Integrate term by term:

\int \frac{1}{2} dx = \frac{x}{2}

\int \frac{1}{2(6x+7)} dx = \frac{1}{2} \cdot \frac{1}{6} \ln|6x+7| = \frac{1}{12} \ln|6x+7|

So the integral is

\frac{x}{2} + \frac{1}{12} \ln|6x+7| + C

Answer:

\frac{x}{2} + \frac{1}{12} \log|6x+7| + C

(Option C).

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