Calculate $\int_e^{e^2}{1+\ln^2 x \over x}dx$

Question

Calculate $$\int_e ^{e^2}{{1+\ln^2x \over x}dx}$$

Personal work:

$$\int_e ^{e^2}{{1+\ln^2x \over x}dx}=\int_e ^{e^2}({1 \over x}+{\ln^2 x \over x})dx=[\ln|x|]_e^{e^2}-\int_e^{e^2}{\ln^2 x \over x}dx=\ln|e^2|-\ln|e|-\int_e^{e^2}{\ln^2 x \over x}dx=|1^2|-|1|-\int_e^{e^2}({1 \over x}*{\ln^2 x}) dx.$$

The antiderivative of $1\over x$ is $\ln|x|$ but for $\ln^2 x$? $${\ln^3 x \over 3} ?$$

Answer 1

Hint: Substitue $$t=\ln(x)$$ then we get $$dt=\frac{1}{x}dx$$

Answer 2

The antiderivative of $1\over x$ is $\ln|x|$ but for $\ln^2 x$? ${\ln^3 x \over 3} ?$

$$\int \ln^2 x dx =x\ln ^2 (x) -\int x 2\ln (x) \frac 1 x dx=x\ln ^2 (x) -2\int \ln (x)dx$$ $$\int \ln (x) dx =x \ln(x)-x$$ For $I=\int \frac {\ln^2 x}x dx$

Substitute $u=\ln(x) \implies dx=xdu$ $$\int \frac {\ln^2 x}x dx=\int u^2du=\frac {\ln(x)^3}{3}$$

Answer 3

$I = \int_e ^{e^2}{{1+\ln^2x \over x}dx}$

let $\ln(x) = t\implies \frac1x\,dx = dt$

$e\rightarrow1\\e^2\rightarrow 2$

$I = \int_1^2(1+t^2\,)dt$

$I = t+\frac{t^3}{3}\bigg|_1^2$

$I =2+\frac{2^3}{3}-1-\frac13$

$I =\frac{10}3$

Answer 4

In $I=\int \dfrac{\ln^2 x dx}{x} $ let $y = \ln x$.

Then $dy = \dfrac{dx}{x}$ so $I=\int y^2 dy =\dfrac{y^3}{3} =\dfrac{\ln^3 x}{3} $.

Answer 5

I think you are on your way with your last conclusion: Let $u=\ln x$, now $du = \frac{1}{x} dx$.

Can you continue now?

Answer 6

As you suggested yourself $$\frac{d}{dx}\left (\frac{\ln(x)^3}{3}\right)=\frac{\ln(x)^2}{x}$$