Solve ∫ x^2/1+m^3 wrt x

$$\int \frac { x ^ { 2 } } { 1 + m ^ { 3 } } d x$$

Evaluate

$$\frac{x^{3}}{3\left(m^{3}+1\right)}+С$$

Solution Steps

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

$$\frac{\int x^{2}\mathrm{d}x}{1+m^{3}}$$

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}.

$$\frac{x^{3}}{3\left(1+m^{3}\right)}$$

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.

$$\frac{x^{3}}{3\left(1+m^{3}\right)}+С$$

Differentiate w.r.t. x

$$\frac{x^{2}}{m^{3}+1}$$