How to solve integration by parts

Question

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So i have this definite integral:

$$\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}x^2 \ln(x^3-3)\,dx.$$

I thought of using integration by parts. So \begin{align} u & =\ln(x^3-3) \\[8pt] du & =\frac{3x^2}{x^3-3} \\[8pt] dv & =x^2 \, dx \\[8pt] v & =\frac{x^3} 3 \end{align}

Answer 1

Integrating by parts: $$\begin{align} \int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}x^2 \ln(x^3-3)\,dx & =\ln(x^3-3)\cdot \frac{x^3}{3}\bigg{|}_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}-\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}\frac{x^3}{3}\cdot \frac{3x^2}{x^3-3}\,dx\\ & =\frac{3+e}{3}-\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}\frac{x^5}{x^3-3}dx=\\ & =\frac{3+e}{3}-\frac13\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}\frac{x^3}{x^3-3} \, d(x^3)\\ & =\frac{3+e}{3}-\frac13\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}\frac{x^3-3+3}{x^3-3} \, d(x^3)\\ & =\frac{3+e}{3}-\frac13\left(x^3+3\ln(x^3-3)\right)\bigg{|}_{\sqrt[3]{4}}^{\sqrt[3]{3+e}} \\ & =\frac{3+e}{3}-\frac13\left(6+e-4\right) \\ & =\frac{1}{3} \end{align}$$

Integration by substitution: $$t=x^3-3 \Rightarrow dt=3x^2dx\\ \int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}x^2 \ln(x^3-3)\,dx=\\ \frac13 \int_1^e \ln t dt=\frac13 (t\ln t|_1^e-\int_1^e t\cdot \frac 1tdt)=\\ \frac13(e-e+1)=\frac13.$$

Answer 2

Hint: Write your integral in the form $$\frac{1}{3}\int 3x^2\ln(x^3-3)dx$$ and Substitute $$t=x^3-3$$

Answer 3

As user Oleg567 suggested in the comments, the $u$-substitution with $u=x^3-3$ works much more efficiently than integration by parts. However, if you're interested in using integration by parts here, I'll provide a sketch of how to use it below.

As you've already written, we'll take

\begin{align*} u &=\ln(x^3-3)\quad&\quad dv&=x^2\,dx\\ du &=\frac{3x^2}{x^3-3}\,dx\quad &\quad v&=\frac13 x^3 \end{align*} The formula for integration by parts is $\int u\,dv=uv-\int v\,du$, so plugging in our parameters above, we have

$$ \int x^2\ln(x^3-3)\,dx=\frac{1}{3}x^3\ln(x^3-3)-\int\frac{x^5}{x^3-3}\,dx. $$

As messy as it is this last integral can be solved by first using long division to get the fraction into a proper fraction form (where the degree of the numerator is less than the degree of the denominator) and then using a $u$-substitution.

Answer 4

I would have done it like this: \begin{align} v & = x^3-3 \\[6pt] dv & = 3x^2 \, dx \\[6pt] \frac{dv} 3 & = x^2 \, dx \end{align} As $x$ goes from $\sqrt[3]4$ to $\sqrt[3]{3+e},$ then $v$ goes from $1$ to $e,$ so we have $$ \int_{\sqrt[3]4}^{\sqrt[3]{3+e}} x^2 \ln(x^3-3) \, dx = \frac 1 3 \int_1^e \ln v\, dv $$ Then I would integrate by parts, as follows: \begin{align} \frac 1 3 \int \ln v \, dv & = \frac 1 3 \int u\, dv & \text{with } u = \ln v \\[10pt] & = \frac 1 3 \left( vu - \int v\,du \right) \\[10pt] & = \frac 1 3 \left( v\ln v - \int v\, \frac{dv} v \right) \\[10pt] & = \frac 1 3 \left( v\ln v - \int dv \right) \\[10pt] & = \frac 1 3 \left[ v\ln v - v \vphantom{\frac 11} \right]_1^e = \frac 1 3. \end{align}

Here's an alternative to integration by parts:

\begin{align} & \int_1^e (\log_e v)\,dv \\[10pt] = {} & \int_1^e \left( \int_1^v \frac 1 w \, dw \right) \, dv \\[10pt] = {} & \iint\limits_{1\,\le\, w\,\le\, v\,\le \,e} \frac 1 w \, d(v,w) \\[10pt] = {} & \int_1^e \left( \int_w^e \frac 1 w \,dv \right)\, dw \\[10pt] = {} & \int_1^e \frac{e-w} w \, dw \\[10pt] = {} & \Big[ e(\log_e w) - w \Big]_1^e \\[10pt] = {} & 1. \end{align}