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3.2(C) · Q99

Q.Evaluate: ∫2x+1x2+4x−5 dx\int \frac{2x+1}{x^2+4x-5}\,dx

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✓ Free question

Derivative of denominator is 2x+42x+4. Write 2x+1=A(2x+4)+B2x+1=A(2x+4)+B: 2A=2⇒A=12A=2\Rightarrow A=1, 4A+B=1⇒B=−34A+B=1\Rightarrow B=-3.

So ∫2x+1x2+4x−5dx=∫2x+4x2+4x−5dx−3∫dxx2+4x−5\displaystyle\int\dfrac{2x+1}{x^2+4x-5}dx=\int\dfrac{2x+4}{x^2+4x-5}dx-3\int\dfrac{dx}{x^2+4x-5}.

First: log⁡∣x2+4x−5∣\log\left|x^2+4x-5\right|.

Second: complete the square, x2+4x−5=(x+2)2−9x^2+4x-5=(x+2)^2-9, giving 16log⁡∣x−1x+5∣\dfrac16\log\left|\dfrac{x-1}{x+5}\right|, so the −3-3 term contributes −12log⁡∣x−1x+5∣-\dfrac12\log\left|\dfrac{x-1}{x+5}\right|.

✓Final answer

log⁡∣x2+4x−5∣−12log⁡∣x−1x+5∣+c\log\left|x^2+4x-5\right|-\dfrac12\log\left|\dfrac{x-1}{x+5}\right|+c

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