Evaluate the integral: I = integral (2+3x)/((3-x)³) dx

Evaluate the integral: $$ I = \int \frac{2+3x}{(3-x)^3}\,dx. $$

Step 1 · Substitute

Let $u=3-x$, so $x=3-u$, $dx=-du$:

$$ 2+3x = 2+3(3-u) = 11-3u. $$

Step 2 · Rewrite the integral

$$ I = \int \frac{11-3u}{u^3}(-du) = -\int \left(11u^{-3}-3u^{-2}\right) du. $$

Step 3 · Integrate term by term

Using $\int u^{-3}du=-\tfrac12u^{-2}$ and $\int u^{-2}du=-u^{-1}$:

$$ I = -\left(11\left(-\frac12u^{-2}\right) - 3\left(-u^{-1}\right)\right) = \frac{11}{2}u^{-2} - 3u^{-1}. $$

Step 4 · Substitute back

With $u=3-x$:

$$ I = \frac{11}{2(3-x)^2} - \frac{3}{3-x} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=3$, where the denominator vanishes.
• This is a rare case where a direct linear substitution outperforms partial fractions, since the denominator is already a single repeated factor and only the numerator needs re-expressing.
• The result can be checked by differentiating $\tfrac{11}{2(3-x)^2}-\tfrac{3}{3-x}$: the first term gives $\tfrac{11}{(3-x)^3}$ (chain rule picks up an extra factor of $-1$ twice, cancelling), and the second gives $\tfrac{-3}{(3-x)^2}=\tfrac{-3(3-x)}{(3-x)^3}=\tfrac{3x-9}{(3-x)^3}$; summing gives $\tfrac{11+3x-9}{(3-x)^3}=\tfrac{2+3x}{(3-x)^3}$, matching the integrand.