Solve ∫ 3x^3(x^5+7)^2 wrt x

$$\int 3 x ^ { 3 } ( x ^ { 5 } + 7 ) ^ { 2 } d x$$

Evaluate

$$\frac{3x^{14}}{14}+\frac{14x^{9}}{3}+\frac{147x^{4}}{4}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{5}+7\right)^{2}.

$$\int 3x^{3}\left(\left(x^{5}\right)^{2}+14x^{5}+49\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 5 and 2 to get 10.

$$\int 3x^{3}\left(x^{10}+14x^{5}+49\right)\mathrm{d}x$$

Use the distributive property to multiply 3x^{3} by x^{10}+14x^{5}+49.

$$\int 3x^{13}+42x^{8}+147x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{13}\mathrm{d}x+\int 42x^{8}\mathrm{d}x+\int 147x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{13}\mathrm{d}x+42\int x^{8}\mathrm{d}x+147\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{13}\mathrm{d}x with \frac{x^{14}}{14}. Multiply 3 times \frac{x^{14}}{14}.

$$\frac{3x^{14}}{14}+42\int x^{8}\mathrm{d}x+147\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{8}\mathrm{d}x with \frac{x^{9}}{9}. Multiply 42 times \frac{x^{9}}{9}.

$$\frac{3x^{14}}{14}+\frac{14x^{9}}{3}+147\int x^{3}\mathrm{d}x$$

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

$$\frac{3x^{14}}{14}+\frac{14x^{9}}{3}+\frac{147x^{4}}{4}$$

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.

$$\frac{3x^{14}}{14}+\frac{14x^{9}}{3}+\frac{147x^{4}}{4}+С$$

Differentiate w.r.t. x

$$3x^{3}\left(x^{5}+7\right)^{2}$$