Solve ∫ (15-6)(5x^2-4x+11)1/2 wrt x

$$\int{ (15-6)(5 { x }^{ 2 } -4x+11) \frac{ 1 }{ 2 } }d x$$

Evaluate

$$\frac{15x^{3}}{2}-9x^{2}+\frac{99x}{2}+С$$

Solution Steps

Subtract 6 from 15 to get 9.

$$\int 9\left(5x^{2}-4x+11\right)\times \frac{1}{2}\mathrm{d}x$$

Multiply 9 and \frac{1}{2} to get \frac{9}{2}.

$$\int \frac{9}{2}\left(5x^{2}-4x+11\right)\mathrm{d}x$$

Use the distributive property to multiply \frac{9}{2} by 5x^{2}-4x+11.

$$\int \frac{45}{2}x^{2}-18x+\frac{99}{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{45x^{2}}{2}\mathrm{d}x+\int -18x\mathrm{d}x+\int \frac{99}{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{45\int x^{2}\mathrm{d}x}{2}-18\int x\mathrm{d}x+\int \frac{99}{2}\mathrm{d}x$$

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{45}{2} times \frac{x^{3}}{3}.

$$\frac{15x^{3}}{2}-18\int x\mathrm{d}x+\int \frac{99}{2}\mathrm{d}x$$

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 -18 times \frac{x^{2}}{2}.

$$\frac{15x^{3}}{2}-9x^{2}+\int \frac{99}{2}\mathrm{d}x$$

Find the integral of \frac{99}{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{15x^{3}}{2}-9x^{2}+\frac{99x}{2}$$

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.

$$\frac{15x^{3}}{2}-9x^{2}+\frac{99x}{2}+С$$

Differentiate w.r.t. x

$$\frac{45x^{2}}{2}-18x+\frac{99}{2}$$