Solve = ∫ d+d/sqrt[3]{a^2+b^2}

$$= \int \frac { d + d } { \sqrt[ 3 ] { a ^ { 2 } + b ^ { 2 } } }$$

Evaluate

$$\frac{2dx}{\sqrt[3]{a^{2}+b^{2}}}+С$$

Solution Steps

Find the integral of \frac{d+d}{\sqrt[3]{a^{2}+b^{2}}} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{d+d}{\sqrt[3]{a^{2}+b^{2}}}x$$

Simplify.

$$\frac{2dx}{\sqrt[3]{a^{2}+b^{2}}}$$

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.

$$\begin{matrix}\frac{2dx}{\sqrt[3]{a^{2}+b^{2}}}+С_{3},&\end{matrix}$$

Differentiate w.r.t. x

$$\frac{2d}{\sqrt[3]{a^{2}+b^{2}}}$$