$$\int_1^e\dfrac{1+x^2\ln x}{x+x^2 \ln x} dx$$


I have tried substitutions like $\ln x = t$, but they are just not helping.

I end up with : $\displaystyle\int_0^1 \dfrac{1+e^{2t}t}{1+ e^t t} dt $

Here method of substitution isn't really possible and integration by parts won't help. How else do I solve it?

  • $\begingroup$ $\ln x=t$ is a good substitution. $\endgroup$
    – Nosrati
    Jul 24 '18 at 2:57

The OP asked, "How else do I solve it?" We proceed to use a straightforward approach that circumvents the use of substitutions.

We need not use any substitutions. Rather, we can write

$$\begin{align} \int_1^e \frac{1+x^2\log(x)}{x+x^2\log(x)}\,dx&=\int_1^e \frac{1-x+x+x^2\log(x)}{x+x^2\log(x)}\,dx\\\\ &=(e-1)+\int_1^e \frac{1-x}{x+x^2\log(x)}\,dx\\\\ &=(e-1)+\int_1^e \left(\frac{1-x}{x+x^2\log(x)}-\frac1x\right)\,dx+\int_1^e \frac1x\,dx\\\\ &=e-\int_1^e \frac{1+\log(x)}{1+x\log(x)}\,dx\\\\ &=e-\left.\left(\log(1+x\log(x)) \right)\right|_1^e\\\\ &=e-\log(1+e) \end{align}$$

  • 2
    $\begingroup$ Hi Mark ! May be, you could be more precise and recall that that $(1+x\log(x))'=1+\log(x)$ $\endgroup$ Jul 24 '18 at 4:37

Let $$I = \int \frac{1+x^2\ln x}{x+x^2\ln x}dx$$

Divide both Numerator and Denominator by $x^2$

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

so $$I = x-\ln \bigg|\frac{1}{x}+\ln x\bigg|+\mathcal{C}.$$

  • $\begingroup$ (+1) Nicely done. $\endgroup$
    – Mark Viola
    Nov 5 '18 at 16:22
  • $\begingroup$ @DXT short and sweet(+1) $\endgroup$
    – Rishi
    Jun 1 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.