Solve ∫ (5+x^2)^3 wrt x

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

Evaluate

$$\frac{x^{7}}{7}+3x^{5}+25x^{3}+125x+С$$

Solution Steps

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

$$\int 125+75x^{2}+15\left(x^{2}\right)^{2}+\left(x^{2}\right)^{3}\mathrm{d}x$$

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

$$\int 125+75x^{2}+15x^{4}+\left(x^{2}\right)^{3}\mathrm{d}x$$

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

$$\int 125+75x^{2}+15x^{4}+x^{6}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 125\mathrm{d}x+\int 75x^{2}\mathrm{d}x+\int 15x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 125\mathrm{d}x+75\int x^{2}\mathrm{d}x+15\int x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

Find the integral of 125 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$125x+75\int x^{2}\mathrm{d}x+15\int x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

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 75 times \frac{x^{3}}{3}.

$$125x+25x^{3}+15\int x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

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 15 times \frac{x^{5}}{5}.

$$125x+25x^{3}+3x^{5}+\int x^{6}\mathrm{d}x$$

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}.

$$125x+25x^{3}+3x^{5}+\frac{x^{7}}{7}$$

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.

$$\frac{x^{7}}{7}+3x^{5}+25x^{3}+125x+С$$

Differentiate w.r.t. x

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