Solve ∫ 8x^6+1/2x^2+1 wrt x

$$\int \frac { 8 x ^ { 6 } + 1 } { 2 x ^ { 2 } + 1 } d x$$

Evaluate

$$\frac{4x^{5}}{5}-\frac{2x^{3}}{3}+x+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{8x^{6}+1}{2x^{2}+1}.

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

Cancel out 2x^{2}+1 in both numerator and denominator.

$$\int 4x^{4}-2x^{2}+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{4}\mathrm{d}x+\int -2x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{4}\mathrm{d}x-2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{4x^{5}}{5}-2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

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 -2 times \frac{x^{3}}{3}.

$$\frac{4x^{5}}{5}-\frac{2x^{3}}{3}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{4x^{5}}{5}-\frac{2x^{3}}{3}+x$$

Simplify.

$$x-\frac{2x^{3}}{3}+\frac{4x^{5}}{5}$$

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.

$$x-\frac{2x^{3}}{3}+\frac{4x^{5}}{5}+С$$

Differentiate w.r.t. x

$$4x^{4}-2x^{2}+1$$