Solve ∫ 10x(5x^2+1) wrt x

$$\int 10 x ( 5 x ^ { 2 } + 1 ) d x$$

Evaluate

$$\frac{25x^{4}}{2}+5x^{2}+С$$

Solution Steps

Use the distributive property to multiply 10x by 5x^{2}+1.

$$\int 50x^{3}+10x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 50x^{3}\mathrm{d}x+\int 10x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$50\int x^{3}\mathrm{d}x+10\int x\mathrm{d}x$$

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

$$\frac{25x^{4}}{2}+10\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 10 times \frac{x^{2}}{2}.

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

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.

$$\frac{25x^{4}}{2}+5x^{2}+С$$

Differentiate w.r.t. x

$$50x^{3}+10x$$