Solve ∫ 10x(5x^2+1)^2 wrt x

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

Evaluate

$$\frac{125x^{6}}{3}+25x^{4}+5x^{2}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x^{2}+1\right)^{2}.

$$\int 10x\left(25\left(x^{2}\right)^{2}+10x^{2}+1\right)\mathrm{d}x$$

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

$$\int 10x\left(25x^{4}+10x^{2}+1\right)\mathrm{d}x$$

Use the distributive property to multiply 10x by 25x^{4}+10x^{2}+1.

$$\int 250x^{5}+100x^{3}+10x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 250x^{5}\mathrm{d}x+\int 100x^{3}\mathrm{d}x+\int 10x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$250\int x^{5}\mathrm{d}x+100\int x^{3}\mathrm{d}x+10\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply 250 times \frac{x^{6}}{6}.

$$\frac{125x^{6}}{3}+100\int x^{3}\mathrm{d}x+10\int x\mathrm{d}x$$

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

$$\frac{125x^{6}}{3}+25x^{4}+10\int x\mathrm{d}x$$

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}.

$$\frac{125x^{6}}{3}+25x^{4}+5x^{2}$$

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.

$$5x^{2}+25x^{4}+\frac{125x^{6}}{3}+С$$

Differentiate w.r.t. x

$$10x\left(5x^{2}+1\right)^{2}$$