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3.2(A) · Q60

Q.Integrate: ∫3e2x+54e2x−5 dx\int \dfrac{3e^{2x}+5}{4e^{2x}-5}\,dx

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Write 3e2x+5=34(4e2x−5)+3543e^{2x}+5=\dfrac34(4e^{2x}-5)+\dfrac{35}{4}.

So the integrand is 34+35/44e2x−5\dfrac34+\dfrac{35/4}{4e^{2x}-5}.

For ∫dx4e2x−5\int\dfrac{dx}{4e^{2x}-5}, let u=e2xu=e^{2x}, du=2u dxdu=2u\,dx: ∫du2u(4u−5)=110ln⁡∣4u−5∣−110ln⁡u\int\dfrac{du}{2u(4u-5)}=\dfrac{1}{10}\ln|4u-5|-\dfrac{1}{10}\ln u, giving 110ln⁡∣4e2x−5∣−x5\dfrac{1}{10}\ln|4e^{2x}-5|-\dfrac x5. …

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