Solve ∫ 3x^2(x^2+5)^2

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

Evaluate

$$\frac{3x^{7}}{7}+6x^{5}+25x^{3}+С$$

Solution Steps

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

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

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

$$\int 3x^{2}\left(x^{4}+10x^{2}+25\right)\mathrm{d}x$$

Use the distributive property to multiply 3x^{2} by x^{4}+10x^{2}+25.

$$\int 3x^{6}+30x^{4}+75x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{6}\mathrm{d}x+\int 30x^{4}\mathrm{d}x+\int 75x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{6}\mathrm{d}x+30\int x^{4}\mathrm{d}x+75\int x^{2}\mathrm{d}x$$

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}. Multiply 3 times \frac{x^{7}}{7}.

$$\frac{3x^{7}}{7}+30\int x^{4}\mathrm{d}x+75\int x^{2}\mathrm{d}x$$

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

$$\frac{3x^{7}}{7}+6x^{5}+75\int x^{2}\mathrm{d}x$$

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}.

$$\frac{3x^{7}}{7}+6x^{5}+25x^{3}$$

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.

$$\frac{3x^{7}}{7}+6x^{5}+25x^{3}+С$$

Differentiate w.r.t. x

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