Evaluate {1}{3x²+13x-10} dx

1/17 \, ≤ft( {3x²-2}{3x²+15} ) + c

1/17 \, ≤ft( {3x-2}{3x²+15} ) + c

1/17 \, ≤ft( 3x+15/3x-2 ) + c

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Integral Calculus

Question : 38 of 73

Marks: +1, -0

Evaluate

\int \frac{1}{3x^{2}+13x-10} dx

Solution:

Concept:

Factor the denominator and use partial fractions to integrate.

Explanation:

The denominator factors as

3x^2 + 13x - 10 = (3x - 2)(x + 5)

Write

\frac{1}{(3x - 2)(x + 5)} = \frac{A}{3x - 2} + \frac{B}{x + 5}

Multiply through:

1 = A(x + 5) + B(3x - 2)

Set

x = \frac{2}{3}

1 = A\left(\frac{17}{3}\right) \Rightarrow A = \frac{3}{17}

Set

x = -5

1 = B(-17) \Rightarrow B = -\frac{1}{17}

Thus

\int \frac{dx}{(3x - 2)(x + 5)} = \frac{3}{17} \int \frac{dx}{3x - 2} - \frac{1}{17} \int \frac{dx}{x + 5}

Integrate:

\frac{3}{17} \cdot \frac{1}{3} \ln|3x - 2| - \frac{1}{17} \ln|x + 5| + C = \frac{1}{17} \ln\left|\frac{3x - 2}{x + 5}\right| + C

Comparing with options: none of A, B, or C match this result (option C gives the negative of the integrand when differentiated).

Answer:

D. None

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