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3.2(A) · Q58

Q.Integrate: ∫4ex−252ex−5 dx\int \dfrac{4e^x-25}{2e^x-5}\,dx

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Write 4ex−25=2(2ex−5)−154e^x-25=2(2e^x-5)-15.

So the integrand is 2−152ex−52-\dfrac{15}{2e^x-5}.

For ∫dx2ex−5\int\dfrac{dx}{2e^x-5}, let u=exu=e^x, dx=du/udx=du/u: ∫duu(2u−5)=15ln⁡∣2u−5∣−15ln⁡u\int\dfrac{du}{u(2u-5)}=\dfrac15\ln|2u-5|-\dfrac15\ln u (partial fractions A=−1/5,B=2/5A=-1/5,B=2/5), giving 15ln⁡∣2ex−5∣−x5\dfrac15\ln|2e^x-5|-\dfrac x5. …

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