MHB Evaluating a definite integral

[Saitama]

Problem:
$$\int_1^e \frac{1+x^2\ln x}{x+x^2\ln x}\,\,dx$$

Attempt:
I tried the substitution $\ln x=t \Rightarrow dx/x=dt$ and got the following integral:
$$\int_0^1 \frac{1+e^{2t}t}{1+e^t t}\,dt$$
I am not sure how to proceed after this.

Any help is appreciated. Thanks!

 

[alyafey22]

Adding and subtracting $x$ might make thinks a little bit easier.

 

[Saitama]

Hi ZaidAlyafey! :)

Adding and subtracting $x$ might make thinks a little bit easier.

I must be missing something but I get stuck on the following:
$$\int_1^e 1+\frac{1-x}{x+x^2\ln x} \,dx$$
How do I proceed after this?

 

[Shobhit]

Write the integrand as
$$\frac{1+x^2 \log(x)}{x+x^2 \log(x)}= 1+\frac{1}{x}-\frac{1+\log(x)}{1+x\log(x)}$$
Now integrate:
$$ \int \frac{1+x^2 \log(x)}{x+x^2 \log(x)}dx=x+\log(x)-\log(1+x\log(x))+C$$

 

[Saitama]

Write the integrand as
$$\frac{1+x^2 \log(x)}{x+x^2 \log(x)}= 1+\frac{1}{x}-\frac{1+\log(x)}{1+x\log(x)}$$

How did you think of that? It doesn't seem obvious to me.

 

[juantheron]

$\displaystyle \int \frac{1+x^2 \ln (x)}{x+x^2\ln(x)}dx$

Now Divide both Numerator and Denominator by $x^2$

$\displaystyle \int\frac{\frac{1}{x^2}+\ln(x)}{\frac{1}{x}+\ln(x)}dx = \int \frac{\left(\frac{1}{x}+\ln(x)\right)-\left(\frac{1}{x}-\frac{1}{x^2}\right)}{\frac{1}{x}+\ln(x)}dx$

Now Let $\displaystyle \frac{1}{x}+\ln(x) = t$, Then $\displaystyle \left(\frac{1}{x}-\frac{1}{x^2}\right)dx = dt$

$\displaystyle = x-\int\frac{1}{t}dt = x-\ln \left|\frac{1}{x}+\ln(x)\right|+\mathbb{C}$

 

[Saitama]

$\displaystyle \int \frac{1+x^2 \ln (x)}{x+x^2\ln(x)}dx$

Now Divide both Numerator and Denominator by $x^2$

$\displaystyle \int\frac{\frac{1}{x^2}+\ln(x)}{\frac{1}{x}+\ln(x)}dx = \int \frac{\left(\frac{1}{x}+\ln(x)\right)-\left(\frac{1}{x}-\frac{1}{x^2}\right)}{\frac{1}{x}+\ln(x)}dx$

Now Let $\displaystyle \frac{1}{x}+\ln(x) = t$, Then $\displaystyle \left(\frac{1}{x}-\frac{1}{x^2}\right)dx = dt$

$\displaystyle = x-\int\frac{1}{t}dt = x-\ln \left|\frac{1}{x}+\ln(x)\right|+\mathbb{C}$

Thank you jacks! That's much easier. :)