Substitution integral with $t=x^3$

Question

$$ \int \frac{3}{t(3+\sqrt[3]{t})} \mathrm{d} t . \text { Substitution } t=x^{3} $$

$$ \int \frac{3}{x^3(3+\sqrt[3]{x^3})}3x^2 $$ $$ \int \frac{9}{x(3+{x})} $$ $$ 9\int \frac{1}{3x+{x^2}} $$

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Now i'm stuck

my calculation so far

Answer 1

You could do partial fraction decomposition, but instead let's multiply the top and bottom by $\frac{1}{x^2}$:

$$3\int\frac{3}{3x+x^2}dx = 3\int \frac{\frac{3}{x^2}}{\frac{3}{x}+1}dx$$

Notice that the numerator is the derivative of the denominator times $-1$, so we have an integral of the form $\int\frac{du}{u}$, which integrates to

$$ -3\log\left(\frac{3}{x}+1\right)+C$$

Then substitute back in for $t$:

$$ -3\log\left(\frac{3}{\sqrt[3]{t}}+1\right)+C$$

Answer 2

$$\int \frac{3}{t(3 + t^{1/3})} \, dt = \int \frac{1}{3 + t^{1/3}} \cdot 9t^{-1/3} \cdot \frac{1}{3} t^{-2/3} \, dt,$$ so with the substitution $$u = 3 + t^{1/3}, \quad du = \frac{1}{3} t^{-2/3} \, dt, \\ t^{-1/3} = \frac{1}{u-3},$$ we obtain $$\int \frac{3}{t(3+t^{1/3})} \, dt = \int \frac{1}{u} \cdot \frac{9}{u-3} \, du = 3 \int \left( \frac{1}{u-3} - \frac{1}{u} \right) \, du = 3 \log\left|\frac{t}{(3+t^{1/3})^3}\right| + C.$$

Answer 3

To finish your calculation:

$$\int\frac{9}{x(3+x)}$$

Take partial fractions

$$=3\int \frac{1}{x+3}dx+2\int\frac{1}{x}dx$$

Let $u=x+3,du=dx$ have

$$=3\int\frac{1}{u}du+2\int\frac{1}{x}dx$$

Then exists $C\in\mathbb{R},s.t.$

$$=3\ln(u)+2\ln(x)+C$$

Substitute $u$ back, we have

$$=3\ln(x+3)+2\ln(x)+C$$