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Exercise 11.2 · Q3

Q.Integrate the following functions with respect to xx:

(i) e3x−6e^{3x-6}
(ii) e8−7xe^{8-7x}
(iii) 16−4x\dfrac{1}{6-4x}
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The exponential and reciprocal-linear (log⁡\log) standard forms both pick up the usual 1/a1/a correction factor for a linear argument.

Step 1. Part (i). a=3a=3; ∫ex dx=ex+c\int e^x\,dx=e^x+c, so

∫e3x−6 dx=13e3x−6+c.\int e^{3x-6}\,dx=\frac13 e^{3x-6}+c.

Step 2. Part (ii). a=−7a=-7; so

∫e8−7x dx=1−7e8−7x+c=−17e8−7x+c.\int e^{8-7x}\,dx=\frac{1}{-7}e^{8-7x}+c=-\frac17 e^{8-7x}+c.

Step 3. Part (iii). a=−4a=-4; ∫1x dx=log⁡∣x∣+c\int \dfrac1x\,dx=\log|x|+c, so

∫dx6−4x=1−4log⁡∣6−4x∣+c=−14log⁡∣6−4x∣+c.\int \frac{dx}{6-4x}=\frac{1}{-4}\log|6-4x|+c=-\frac14\log|6-4x|+c. …

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