Solving the the Integral: $\int\frac{\ln(x+3)}{(x+3)^{\frac{1}{3}}}dx$

Question

I am attempting to solve the following integral:

$$\int\frac{\ln(x+3)}{(x+3)^{\frac{1}{3}}}dx$$

I have tried the following but I am not sure if it is correct:

$u = x+3$

$\frac{du}{dx} = 1 \Rightarrow du = dx$

Therefore:

$$\int\frac{\ln(x+3)}{(x+3)^{\large\frac{1}{3}}}dx = \int\frac{\ln(u)}{u^{\large\frac{1}{3}}}du$$

I am not sure where to go from here.

Answer 1

setting $$u^{1/3}=t$$ then we have $$u=t^3$$ and $$du=3t^2dt$$ and we have $$\int\frac{\ln(t^3)}{t}3t^2dt$$ the result is given by $$3/2\,\ln \left( x+3 \right) \left( x+3 \right) ^{2/3}-9/4\, \left( x +3 \right) ^{2/3} $$

Answer 2

A single substitution followed by integration by parts will work: let $$v = (x+3)^{2/3}, \quad dv = \frac{2}{3}(x+3)^{-1/3} \, dx.$$ Then $x+3 = v^{3/2}$ and we obtain $$\int \frac{\log (x+3)}{(x+3)^{1/3}} \, dx = \int \frac{3}{2} \log v^{3/2} \, dv = \frac{9}{4} \int \log v \, dv.$$ This last integral is easily computed by parts.

Answer 3

Let $I(a) =\int \ln(t)t^a dt $.

Based on the other answers, I'll try to find a substitution $t = x^b$ that will make the integral easy to evaluate.

If $t = x^b$, then $dt = bx^{b-1}dx $.

Therefore

$\begin{array}\\ I(a) &=\int \ln(x^b)(x^b)^a bx^{b-1}dx\\ &=b^2\int \ln(x)x^{ab+b-1}dx\\ \end{array} $

If we choose $b$ so that $ab+b-1 = 0$, then $I(a) =b^2\int \ln(x) dx =b^2(x \ln x - x) $. This requires $b = \frac1{a+1} $ which can be done except when $a = -1$.

When $a = -1$, $I(a) =\int \dfrac{\ln t dt}{t} =\dfrac{\ln^2(t)}{2} $.

Therefore, except when $a = -1$, since $x = t^{1/b} = t^{a+1} $,

$\begin{array}\\ I(a) &=b^2(x \ln x - x)\\ &=\dfrac1{(a+1)^2}(t^{a+1} \ln t^{a+1} - t^{a+1})\\ &=\dfrac{t^{a+1}}{(a+1)^2}((a+1) \ln t - 1)\\ &=\dfrac{t^{a+1}\ln t}{a+1}-\dfrac{t^{a+1}}{(a+1)^2}\\ \text{Check:}\\ I'(a) &=\dfrac1{(a+1)^2}\left(t^{a+1}\dfrac{a+1}{t}+(a+1)t^a((a+1) \ln t - 1)\right)\\ &=\dfrac1{(a+1)^2}\left((a+1)t^{a}+(a+1)^2t^a \ln t - (a+1)t^a\right)\\ &=t^a \ln t\\ \end{array} $

When $a = -\frac13$, $I(-\frac13) =\dfrac{t^{a+1}\ln t}{a+1}-\dfrac{t^{a+1}}{(a+1)^2} =\dfrac{t^{2/3}\ln t}{2/3}-\dfrac{t^{2/3}}{(2/3)^2} =\dfrac{3t^{2/3}\ln t}{2}-\dfrac{9t^{2/3}}{4} $.

Answer 4

Another way: $$\dfrac{d}{d\alpha} t^\alpha = t^\alpha \log(t)$$

So if $$J(\alpha) = \int (x+3)^\alpha \; dx = \frac{(x+3)^{\alpha + 1}}{\alpha + 1} + C(\alpha) $$ your integral is $$ J'(-1/3) = \frac{3}{2} (x+3)^{2/3} \ln(x+3) - \frac{9}{4} (x+3)^{2/3} + C$$