Solve ∫ (from - 1 to 1) of ((2x^2-3)^2+6x^4) wrt x

$$\int _ { - 1 } ^ { 1 } ( ( 2 x ^ { 2 } - 3 ) ^ { 2 } + 6 x ^ { 4 } ) d x$$

Evaluate

$$14$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x^{2}-3\right)^{2}.

$$\int _{-1}^{1}4\left(x^{2}\right)^{2}-12x^{2}+9+6x^{4}\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{-1}^{1}4x^{4}-12x^{2}+9+6x^{4}\mathrm{d}x$$

Combine 4x^{4} and 6x^{4} to get 10x^{4}.

$$\int _{-1}^{1}10x^{4}-12x^{2}+9\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 10x^{4}-12x^{2}+9\mathrm{d}x$$

Integrate the sum term by term.

$$\int 10x^{4}\mathrm{d}x+\int -12x^{2}\mathrm{d}x+\int 9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$10\int x^{4}\mathrm{d}x-12\int x^{2}\mathrm{d}x+\int 9\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 10 times \frac{x^{5}}{5}.

$$2x^{5}-12\int x^{2}\mathrm{d}x+\int 9\mathrm{d}x$$

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 -12 times \frac{x^{3}}{3}.

$$2x^{5}-4x^{3}+\int 9\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{5}-4x^{3}+9x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$9\times 1-4\times 1^{3}+2\times 1^{5}-\left(9\left(-1\right)-4\left(-1\right)^{3}+2\left(-1\right)^{5}\right)$$

Simplify.

$$14$$