Using u-substitution on an integral

Question

I have the integral of $$\int_{1}^{3} \dfrac{x^4}{\ln(x)+2}{dx}$$

I don't want to evaluate it, but I'm supposed to simply use u-substitution and let $u = \ln(x)+2$. The problem is that the correct answer is supposed to give you $\int_{2}^{ln(3)+2} \dfrac{e^{5u-10}}{u}{du}$

Now, I do understand everything up to the part of $e^{5u-10}$. When I take $u - 2 = ln(x)$ and take the equivalent of $x^4$ which is $e^{4\ln(x)}$, I get $e^{4(u-2)}$, which is then obviously $e^{4u-8}$ and not $e^{5u-10}$. Is the exercise faulty or am I making some mistake I'm not aware of?

Answer 1

If $u=\ln(x)+2$, then $x=e^{u-2}$ and $\mathrm dx=e^{u-2}\,\mathrm du$. The integral then becomes

$$\int_1^3 \frac{x^4}{\ln(x)+2}\,\mathrm dx=\int_2^{\ln(3)+2}\frac{e^{4(u-2)}}u(e^{u-2}\,\mathrm du)=\int_2^{\ln(3)+2}\frac{e^{5u-10}}u\,\mathrm du$$

The missing link between this and your solution is accounting for how $\mathrm dx$ is transformed.

Answer 2

If you take $u = \ln(x) +2$, then notice that $du = 1/x \cdot dx$. But, from the first, it follows that $x = e^{u-2}$. Thus, solving for $dx = x \cdot du$ and using that, we have

$$\int \frac{x^4}{\ln(x) + 2} dx = \int \frac{(e^{u-2})^4}{u} \cdot e^{u-2} du$$