Solve ∫ (x^3+5)^2 wrt x

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

Evaluate

$$\frac{x^{7}}{7}+\frac{5x^{4}}{2}+25x+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{3}+5\right)^{2}.

$$\int \left(x^{3}\right)^{2}+10x^{3}+25\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\int x^{6}+10x^{3}+25\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{6}\mathrm{d}x+\int 10x^{3}\mathrm{d}x+\int 25\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{6}\mathrm{d}x+10\int x^{3}\mathrm{d}x+\int 25\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}.

$$\frac{x^{7}}{7}+10\int x^{3}\mathrm{d}x+\int 25\mathrm{d}x$$

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

$$\frac{x^{7}}{7}+\frac{5x^{4}}{2}+\int 25\mathrm{d}x$$

Find the integral of 25 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{7}}{7}+\frac{5x^{4}}{2}+25x$$

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.

$$25x+\frac{5x^{4}}{2}+\frac{x^{7}}{7}+С$$

Differentiate w.r.t. x

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