Solve ∫ 5v(v^2+2)^2dv

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

Evaluate

$$\frac{5v^{6}}{6}+5v^{4}+10v^{2}+С$$

Solution Steps

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

$$\int 5v\left(\left(v^{2}\right)^{2}+4v^{2}+4\right)\mathrm{d}v$$

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

$$\int 5v\left(v^{4}+4v^{2}+4\right)\mathrm{d}v$$

Use the distributive property to multiply 5v by v^{4}+4v^{2}+4.

$$\int 5v^{5}+20v^{3}+20v\mathrm{d}v$$

Integrate the sum term by term.

$$\int 5v^{5}\mathrm{d}v+\int 20v^{3}\mathrm{d}v+\int 20v\mathrm{d}v$$

Factor out the constant in each of the terms.

$$5\int v^{5}\mathrm{d}v+20\int v^{3}\mathrm{d}v+20\int v\mathrm{d}v$$

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

$$\frac{5v^{6}}{6}+20\int v^{3}\mathrm{d}v+20\int v\mathrm{d}v$$

Since \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} for k\neq -1, replace \int v^{3}\mathrm{d}v with \frac{v^{4}}{4}. Multiply 20 times \frac{v^{4}}{4}.

$$\frac{5v^{6}}{6}+5v^{4}+20\int v\mathrm{d}v$$

Since \int v^{k}\mathrm{d}v=\frac{v^{k+1}}{k+1} for k\neq -1, replace \int v\mathrm{d}v with \frac{v^{2}}{2}. Multiply 20 times \frac{v^{2}}{2}.

$$\frac{5v^{6}}{6}+5v^{4}+10v^{2}$$

If F\left(v\right) is an antiderivative of f\left(v\right), then the set of all antiderivatives of f\left(v\right) is given by F\left(v\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{5v^{6}}{6}+5v^{4}+10v^{2}+С$$

Differentiate w.r.t. v

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