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Exercise 11.5 · Q14

Q.(3x+4)3x+7(3x+4)\sqrt{3x+7}

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Re-expressing the linear factor in terms of (3x+7)(3x+7) turns the product into a difference of two power-rule integrals.

Step 1. Rewrite the linear factor. 3x+4=(3x+7)−33x+4=(3x+7)-3, so (3x+4)3x+7=(3x+7)3/2−3(3x+7)1/2(3x+4)\sqrt{3x+7}=(3x+7)^{3/2}-3(3x+7)^{1/2}.

Step 2. Substitute u=3x+7u=3x+7, du=3 dxdu=3\,dx. ∫(3x+7)3/2dx=13∫u3/2du=13⋅25u5/2=215u5/2\displaystyle\int(3x+7)^{3/2}dx=\dfrac13\int u^{3/2}du=\dfrac13\cdot\dfrac{2}{5}u^{5/2}=\dfrac{2}{15}u^{5/2}, and ∫3(3x+7)1/2dx=3⋅13∫u1/2du=23u3/2\displaystyle\int3(3x+7)^{1/2}dx=3\cdot\dfrac13\int u^{1/2}du=\dfrac23u^{3/2}.

Step 3. Combine and re-substitute. ∫(3x+4)3x+7 dx=215(3x+7)5/2−23(3x+7)3/2+c\displaystyle\int(3x+4)\sqrt{3x+7}\,dx=\dfrac{2}{15}(3x+7)^{5/2}-\dfrac23(3x+7)^{3/2}+c. …

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