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Chapter 12: Problem 17

Use partial fractions to find the indefinite integral. $$ \int \frac{1}{2 x^{2}-x} d x $$

Short Answer

Expert verified
So the indefinite integral of \( \frac{1}{2x^{2} - x} \) is -2ln|x| - \(\frac{1}{2x}\) + C, where C is the constant of integration.

Step by step solution

01

Rewrite the original function

Split the fraction into partial fractions, recall that \( \frac{1}{ax^{2}-b} = \frac{A}{x} + \frac{B}{x^{2}} \). Now, equate this with the original fraction to find the coefficients. So: \( \frac{1}{2x^2 - x} = \frac{A}{x} + \frac{B}{2x^2} \). Multiplying through by the common denominator gives 1 = A(2x) + B. Setting x = 0 gives B = 1 and substituting B = 1 into the original equation gives A = -2.
02

Express the updated function

The function can be written as: \( \frac{-2}{x} + \frac{1}{2x^{2}} \)
03

Perform the integration

Now, perform the integration on each part. The integral of \( \frac{-2}{x} \) is -2ln|x| and for \( \frac{1}{2x^{2}} \), it is \( -\frac{1}{2x} \)

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