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3.2(C) · Q106

Q.Evaluate: ∫e3x−e2xex+1 dx\int \frac{e^{3x}-e^{2x}}{e^x+1}\,dx

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Substitute t=ex, dt=ex dx⇒dx=dttt=e^x,\ dt=e^x\,dx\Rightarrow dx=\dfrac{dt}{t}. Then e3x=t3, e2x=t2e^{3x}=t^3,\ e^{2x}=t^2, so the integrand becomes

t3−t2t+1⋅1t=t2−tt+1\dfrac{t^3-t^2}{t+1}\cdot\dfrac1t=\dfrac{t^2-t}{t+1}.

Polynomial division: t2−t=(t+1)(t−2)+2t^2-t=(t+1)(t-2)+2, so t2−tt+1=t−2+2t+1\dfrac{t^2-t}{t+1}=t-2+\dfrac{2}{t+1}. …

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