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Worked solution

Match exponential derivatives and bases

Integrate with respect to xx: (a) 2e2x−12e^{2x-1}; (b) 3xex2−13xe^{x^2-1}; (c) ex−1ex\sqrt{e^x}-\frac1{\sqrt{e^x}}; (d) esin⁡−1x1−x2\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}; (e) x351−x4x^3 5^{1-x^4}.

CJC · Lecture · Example 4

(a)

With f(x)=2x−1f(x)=2x-1 and f′(x)=2f^{\prime}(x)=2, ∫2e2x−1 dx=e2x−1+C\int2e^{2x-1}\,dx=e^{2x-1}+C.

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(b)

With f(x)=x2−1f(x)=x^2-1, f′(x)=2xf^{\prime}(x)=2x, ∫3xex2−1 dx=32∫2xex2−1 dx=32ex2−1+C\int3xe^{x^2-1}\,dx=\frac32\int2xe^{x^2-1}\,dx=\frac32e^{x^2-1}+C.

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(c)

Rewrite the radicals as exponentials: ex−1/ex=ex/2−e−x/2\sqrt{e^x}-1/\sqrt{e^x}=e^{x/2}-e^{-x/2}. The Teacher names f=x/2f=x/2, g=−x/2g=-x/2, with derivatives 1/2,−1/21/2,-1/2. Thus ∫(ex/2−e−x/2) dx=2∫12ex/2 dx−(−2)∫(−12)e−x/2 dx=2ex/2+2e−x/2+C\int(e^{x/2}-e^{-x/2})\,dx=2\int\frac12e^{x/2}\,dx-(-2)\int(-\frac12)e^{-x/2}\,dx=2e^{x/2}+2e^{-x/2}+C.

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(d)

Here sin⁡−1x\sin^{-1}x means arcsine, not reciprocal sine. Take f(x)=sin⁡−1xf(x)=\sin^{-1}x with f′(x)=1/1−x2f^{\prime}(x)=1/\sqrt{1-x^2}. Therefore ∫esin⁡−1x1−x2 dx=esin⁡−1x+C\int\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}\,dx=e^{\sin^{-1}x}+C for −1<x<1-1<x<1.

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(e)

With f(x)=1−x4f(x)=1-x^4 and f′(x)=−4x3f^{\prime}(x)=-4x^3, ∫x351−x4 dx=−14∫(−4x3)51−x4 dx=−1451−x4ln⁡5+C\int x^3 5^{1-x^4}\,dx=-\frac14\int(-4x^3)5^{1-x^4}\,dx=-\frac14\frac{5^{1-x^4}}{\ln5}+C.

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Common mistakes

In (c), the negative exponent and subtraction give a positive second antiderivative term. In (e), retain both the negative quarter and division by ln 5.

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