AHL 5.16—Integration by substitution, parts and repeated parts

The integrand is a product of a Logarithmic and an Algebraic function. By LIATE, logarithmic comes first, so let $u=\ln x$ and $\frac{dv}{dx}=x^3$.

Differentiating: $\frac{du}{dx}=\frac{1}{x}$. Integrating: $v=\frac{x^4}{4}$.

$\int x^3\ln xdx=\frac{x^4}{4}\ln x-\int \frac{x^4}{4}\cdot \frac{1}{x}dx=\frac{x^4\ln x}{4}-\int \frac{x^3}{4}dx$

$\int \frac{x^3}{4}dx=\frac{1}{4}\cdot \frac{x^4}{4}=\frac{x^4}{16}$

$\int x^3\ln xdx=\frac{x^4\ln x}{4}-\frac{x^4}{16}+C$

This requires integration by parts with $u=\ln x$. After applying the formula we expect a term $\frac{x^4\ln x}{4}$ minus a pure polynomial integral.

The option $\frac{x^4\ln x}{4}+\frac{x^4}{16}+C$ has a plus sign before $\frac{x^4}{16}$. Differentiating this gives $x^3\ln x+\frac{x^3}{4}+\frac{x^3}{4}=x^3\ln x+\frac{x^3}{2}$, which does not equal $x^3\ln x$. Eliminate.

The option $\frac{x^4\ln x}{4}-\frac{x^4}{4}+C$: differentiating gives $x^3\ln x+\frac{x^3}{4}-x^3=x^3\ln x-\frac{3x^3}{4}\ne x^3\ln x$. Eliminate.

The option $x^4\ln x-\frac{x^4}{4}+C$: differentiating gives $4x^3\ln x+x^3-x^3=4x^3\ln x\ne x^3\ln x$. Eliminate.

Only $\frac{x^4\ln x}{4}-\frac{x^4}{16}+C$ remains. Differentiating confirms: $x^3\ln x+\frac{x^3}{4}-\frac{x^3}{4}=x^3\ln x$. ✓

By LIATE: let $u=\ln x$, $\frac{dv}{dx}=x^5$. Then $\frac{du}{dx}=\frac{1}{x}$ and $v=\frac{x^6}{6}$.

$\int x^5\ln xdx=\frac{x^6\ln x}{6}-\int \frac{x^6}{6}\cdot \frac{1}{x}dx=\frac{x^6\ln x}{6}-\frac{1}{6}\int x^{5}dx=\frac{x^6\ln x}{6}-\frac{x^6}{36}$

${\left[\frac{x^6\ln x}{6}-\frac{x^6}{36}\right]}_1^e=\left(\frac{e^6\cdot 1}{6}-\frac{e^6}{36}\right)-\left(\frac{1\cdot 0}{6}-\frac{1}{36}\right)$

$=\frac{e^6}{6}-\frac{e^6}{36}+\frac{1}{36}=\frac{e^6}{6}-\frac{e^6-1}{36}$

The answer is $\frac{e^6}{6}-\frac{e^6-1}{36}$, noting that $\frac{e^6}{6}-\frac{e^6}{36}+\frac{1}{36}=\frac{e^6}{6}-\frac{e^6-1}{36}$. Both expressions are equivalent.

Integration by parts on ${\int }_1^e{x}^5\ln xdx$ gives ${\left[\frac{x^6\ln x}{6}-\frac{x^6}{36}\right]}_1^e$. At $x=1$: $\ln 1=0$ so the term is $-\frac{1}{36}$. At $x=e$: $\ln e=1$ so the term is $\frac{e^6}{6}-\frac{e^6}{36}$.

The option $\frac{e^6}{6}-\frac{e^6+1}{36}$ would require a $-\frac{1}{36}$ contribution from the lower limit but with a negative sign overall, giving $\frac{e^6}{6}-\frac{e^6}{36}-\frac{1}{36}$, which contradicts our calculation. Eliminate.

The option $\frac{e^6}{36}+\frac{1}{36}$ lacks the $\frac{e^6}{6}$ term entirely. This is clearly wrong since the antiderivative evaluated at $e$ includes $\frac{e^6}{6}$. Eliminate.

Both $\frac{e^6}{6}-\frac{e^6-1}{36}$ and $\frac{e^6}{6}-\frac{e^6}{36}+\frac{1}{36}$ are equivalent expressions: $\frac{e^6-1}{36}=\frac{e^6}{36}-\frac{1}{36}$, so they are the same value. The answer listed as $\frac{e^6}{6}-\frac{e^6-1}{36}$ is the correct simplified form.

The correct answer is $\frac{e^6}{6}-\frac{e^6-1}{36}$, which equals $\frac{6e^6-e^6+1}{36}=\frac{5e^6+1}{36}$. This is consistent with our computation.