Solve ∫ intxe^xz wrt x

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

Evaluate

$$\left\{\begin{matrix}\frac{xze^{xz}-2e^{xz}}{z^{3}}+С\left(x+1\right),&z\neq 0\\Сx+\frac{x^{3}}{6}+С_{1},&z=0\end{matrix}\right.$$

Solution Steps

Simplify.

$$\left(\frac{-e^{xz}+xze^{xz}}{z^{2}}+С_{3}\right)x-\frac{x^{2}e^{xz}}{z}+\frac{2\left(-e^{xz}+xze^{xz}\right)}{z^{3}}$$

Integrate the sum term by term.

$$\int \frac{x^{2}}{2}\mathrm{d}x+\int С_{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{\int x^{2}\mathrm{d}x}{2}+\int С_{4}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply \frac{1}{2} times \frac{x^{3}}{3}.

$$\frac{x^{3}}{6}+\int С_{4}\mathrm{d}x$$

Find the integral of С_{4} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{6}+С_{4}x$$

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.

$$\left\{\begin{matrix}\left(\frac{-e^{xz}+xze^{xz}}{z^{2}}+С_{3}\right)x-\frac{x^{2}e^{xz}}{z}+\frac{2\left(-e^{xz}+xze^{xz}\right)}{z^{3}}+С_{7},&\\\frac{x^{3}}{6}+С_{4}x+С_{7},&\end{matrix}\right.$$

Differentiate w.r.t. x

$$\left\{\begin{matrix}\frac{xze^{xz}-e^{xz}}{z^{2}}+С,&z\neq 0\\\frac{x^{2}}{2}+С,&z=0\end{matrix}\right.$$