Solve ∫ 3x-2x/frac{9{3x wrt x

$$\int{ \frac{ 3x-2x }{ \frac{ 9 }{ 3x } } }d x$$

Evaluate

$$\frac{x^{3}}{9}+С$$

Solution Steps

Combine 3x and -2x to get x.

$$\int \frac{x}{\frac{9}{3x}}\mathrm{d}x$$

Divide x by \frac{9}{3x} by multiplying x by the reciprocal of \frac{9}{3x}.

$$\int \frac{x\times 3x}{9}\mathrm{d}x$$

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

$$\int \frac{x^{2}\times 3}{9}\mathrm{d}x$$

Divide x^{2}\times 3 by 9 to get x^{2}\times \frac{1}{3}.

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

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

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

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}.

$$\frac{x^{3}}{9}$$

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.

$$\frac{x^{3}}{9}+С$$

Differentiate w.r.t. x

$$\frac{x^{2}}{3}$$