Solve ∫ (2-x^2)(7x-3/5) wrt x=

$$\int ( 2 - x ^ { 2 } ) ( 7 x - 3 / 5 ) d x =$$

Evaluate

$$-\frac{7x^{4}}{4}+\frac{x^{3}}{5}+7x^{2}-\frac{6x}{5}+С$$

Solution Steps

Use the distributive property to multiply 2-x^{2} by 7x-\frac{3}{5}.

$$\int 14x-\frac{6}{5}-7x^{3}+\frac{3}{5}x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 14x\mathrm{d}x+\int -\frac{6}{5}\mathrm{d}x+\int -7x^{3}\mathrm{d}x+\int \frac{3x^{2}}{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$14\int x\mathrm{d}x+\int -\frac{6}{5}\mathrm{d}x-7\int x^{3}\mathrm{d}x+\frac{3\int x^{2}\mathrm{d}x}{5}$$

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}. Multiply 14 times \frac{x^{2}}{2}.

$$7x^{2}+\int -\frac{6}{5}\mathrm{d}x-7\int x^{3}\mathrm{d}x+\frac{3\int x^{2}\mathrm{d}x}{5}$$

Find the integral of -\frac{6}{5} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$7x^{2}-\frac{6x}{5}-7\int x^{3}\mathrm{d}x+\frac{3\int x^{2}\mathrm{d}x}{5}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -7 times \frac{x^{4}}{4}.

$$7x^{2}-\frac{6x}{5}-\frac{7x^{4}}{4}+\frac{3\int x^{2}\mathrm{d}x}{5}$$

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{3}{5} times \frac{x^{3}}{3}.

$$7x^{2}-\frac{6x}{5}-\frac{7x^{4}}{4}+\frac{x^{3}}{5}$$

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.

$$7x^{2}-\frac{6x}{5}-\frac{7x^{4}}{4}+\frac{x^{3}}{5}+С$$

Differentiate w.r.t. x

$$\frac{\left(3-35x\right)\left(x^{2}-2\right)}{5}$$