Solve ∫ 1/x*1/3x^2 wrt x

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

Evaluate

$$\frac{x^{2}}{6}+С$$

Solution Steps

Multiply \frac{1}{x} times \frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

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

Express \frac{1}{x\times 3}x^{2} as a single fraction.

$$\int \frac{x^{2}}{x\times 3}\mathrm{d}x$$

Cancel out x in both numerator and denominator.

$$\int \frac{x}{3}\mathrm{d}x$$

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

$$\frac{\int x\mathrm{d}x}{3}$$

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

$$\frac{x^{2}}{6}$$

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.

$$\frac{x^{2}}{6}+С$$

Differentiate w.r.t. x

$$\frac{x}{3}$$