How should I evaluate this indefinite integral involving natural logarithm?

Question

Find $\int \ln(5x^3)dx$.

Based on my Ti-89 graphing calculator, the answer I found is $x\cdot \ln(x^3)+(\ln(5)-3)\cdot x$. But I want to know how to get to this answer. I know that by the general formula/rule, $\int \ln(x)dx=\frac{1}{x}+c$ but how to apply that to this problem?

Answer 1

Remember these two rules: $$\ln(ab)=\ln(a)+\ln(b)\text{ and }\ln(a^b)=b\ln(a)$$So the integral really is $$\int(\ln5+3\ln x)dx$$Can you continue from here?

Edit: Use that $\int\ln xdx=x(\ln x-1)+C$, what you thought was the integral is actually its derivative with a $+C$ mixed in.

Answer 2

Your statement $$\int \ln x \, dx = \frac{1}{x} + C$$ is incorrect. You've mixed up integration and differentiation. The correct formula is $$\frac{d}{dx}\bigl[ \ln x \bigr] = \frac{1}{x},$$ and $$\int \frac{1}{x} \, dx = \ln |x| + C.$$

Instead, you would use integration by parts with the choice $$u = \ln x, \quad du = \frac{1}{x} \, dx, \\ dv = dx, \quad v = x,$$ to obtain $$\int \ln x \, dx = uv - \int v \, du = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C.$$

Answer 3

Combining what others have noted, this would be the full solution to your integral:

From the property $\ln(ab)=\ln a+\ln b$, we rewrite the integrand $\ln(5x^3)=\ln5+\ln x^3$, so: $$\int \ln(5x^3)dx=\int \ln5+\ln x^3dx=\int \ln5dx+\int\ln x^3dx=x\ln5+\int \ln x^3dx$$ From the property $\ln a^b=b\ln a$, we rewrite $\ln x^3=3\ln x$, therefore: $$x\ln5+3\int1\cdot \ln xdx$$ We can find the integral of $\ln x$ using integration by parts (your formula is almost correct, except it works the other way around, since it applies to the derivative of $\ln x$, not its integral): $$u=\color{#C00000}{\ln x} \quad du=\color{#008080}{\frac{1}{x}}dx$$ $$dv=1 dx \quad v=\color{#C00000}{x}$$ $$\int\ln xdx=\color{#C00000}{x\ln x}-\int \color{#C00000}{x}\cdot\ \color{#008080}{\frac{1}{x}} dx=x\ln x-x+c_1$$ Using our previous results, we find the initial integral: $$x\ln5+3x\ln x-3x+c$$ We can again use the property $b\ln a=\ln a^b$ to rewrite $3\ln x=\ln x^3$. Thus, the final answer: $$\int \ln(5x^3)dx=x\ln x^3+x\ln5-3x+c$$ Which is exactly what you found using a calculator, if you factor out $x$ for the terms $x\ln5$ and $3x$.
Hope it helps :)