Solve ∫ lnxe^lnx wrt x

$$\int \ln x e ^ { \ln x } d x$$

Evaluate

$$2x\ln(x)-2x+С$$

Solution Steps

Apply rule a^{\log_{a}\left(b\right)}=b, where a=e and b=x.

$$\int \log_{e}\left(xx\right)\mathrm{d}x$$

Multiply x and x to get x^{2}.

$$\int \log_{e}\left(x^{2}\right)\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$\frac{\ln(x^{2})x-2x}{\ln(e)}$$

Simplify.

$$\ln(x^{2})x-2x$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\ln(x^{2})x-2x+С$$

Differentiate w.r.t. x

$$2\ln(x)$$