Solve ∫ (x^2+3)^22x wrt x

$$\int ( x ^ { 2 } + 3 ) ^ { 2 } 2 x d x$$

Evaluate

$$\frac{x^{6}}{3}+3x^{4}+9x^{2}+С$$

Solution Steps

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

$$\int \left(\left(x^{2}\right)^{2}+6x^{2}+9\right)\times 2x\mathrm{d}x$$

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

$$\int \left(x^{4}+6x^{2}+9\right)\times 2x\mathrm{d}x$$

Use the distributive property to multiply x^{4}+6x^{2}+9 by 2.

$$\int \left(2x^{4}+12x^{2}+18\right)x\mathrm{d}x$$

Use the distributive property to multiply 2x^{4}+12x^{2}+18 by x.

$$\int 2x^{5}+12x^{3}+18x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{5}\mathrm{d}x+\int 12x^{3}\mathrm{d}x+\int 18x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{5}\mathrm{d}x+12\int x^{3}\mathrm{d}x+18\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply 2 times \frac{x^{6}}{6}.

$$\frac{x^{6}}{3}+12\int x^{3}\mathrm{d}x+18\int x\mathrm{d}x$$

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 12 times \frac{x^{4}}{4}.

$$\frac{x^{6}}{3}+3x^{4}+18\int x\mathrm{d}x$$

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}.

$$\frac{x^{6}}{3}+3x^{4}+9x^{2}$$

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.

$$9x^{2}+3x^{4}+\frac{x^{6}}{3}+С$$

Differentiate w.r.t. x

$$2x\left(x^{2}+3\right)^{2}$$