Find the integral from 0 to 1 of (2x+3)/(6x+7)3

Evaluate the following integral:

$\int_0^1 \frac{2x+3}{(6x+7)^3} \, dx.$

First we need to get this into a form we can do something with (if you have a more elegant solution, please let us know):

$\begin{align*} \int_0^1 \frac{2x+3}{(6x+7)^3} \, dx &= \frac{1}{3} \int_0^1 \frac{6x + 9}{(6x+7)^3} \, dx \\ &= \frac{1}{3} \int_0^1 \frac{6x+7}{(6x+7)^3} \, dx + \frac{1}{3} \int_0^1 \frac{2}{(6x+7)^3} \, dx \\ &= \frac{1}{3} \int_0^1 \frac{1}{(6x+7)^2} \, dx + \frac{1}{9} \int_0^1 \frac{6}{(6x+7)^3} \, dx \\ &= \frac{1}{18} \int_0^1 \frac{6}{(6x+7)^2} \, dx + \frac{1}{9} \int_0^1 \frac{6}{(6x+7)^3} \, dx. \end{align*}$

Now, in both integrals we make the substitution,

$u = 6x+7 \qquad \implies \qquad du = 6 \, dx.$

Which gives us new bounds of integration,

$u(0) = 7 \qquad \text{and} \qquad u(1) = 13.$

Therefore,

$\begin{align*} \int_0^1 \frac{2x+3}{(6x+7)^3} \, dx &= \frac{1}{18} \int_7^{13} \frac{1}{u^2} \, du + \frac{1}{9} \int_7^{13} \frac{1}{u^3} \, du \\ &= \frac{1}{18} \left( -\frac{1}{u} \Big \rvert_7^{13} \right) + \frac{1}{18} \left( -\frac{1}{u^2} \Big \rvert_7^{13} \right) \\ &= \frac{1}{18} \left( - \frac{u+1}{u^2} \Big \rvert_7^{13} \right) \\ &= \frac{1}{18} \left( \frac{8}{49} - \frac{14}{169} \right) \\ &= \frac{1}{18} \left( \frac{666}{8281} \right) \\ &= \frac{37}{8281}. \end{align*}$

Stumbling Robot

Find the integral from 0 to 1 of (2x+3)/(6x+7)³

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