Solve ∫ 36x^4-109x^2+25/3x-5 wrt x

$$\int{ \frac{ 36 { x }^{ 4 } -109 { x }^{ 2 } +25 }{ 3x-5 } }d x$$

Evaluate

$$3x^{4}+\frac{20x^{3}}{3}-\frac{3x^{2}}{2}-5x+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{36x^{4}-109x^{2}+25}{3x-5}.

$$\int \frac{\left(2x-1\right)\left(3x-5\right)\left(2x+1\right)\left(3x+5\right)}{3x-5}\mathrm{d}x$$

Cancel out 3x-5 in both numerator and denominator.

$$\int \left(2x-1\right)\left(2x+1\right)\left(3x+5\right)\mathrm{d}x$$

Expand the expression.

$$\int 12x^{3}+20x^{2}-3x-5\mathrm{d}x$$

Integrate the sum term by term.

$$\int 12x^{3}\mathrm{d}x+\int 20x^{2}\mathrm{d}x+\int -3x\mathrm{d}x+\int -5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$12\int x^{3}\mathrm{d}x+20\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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

$$3x^{4}+20\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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 20 times \frac{x^{3}}{3}.

$$3x^{4}+\frac{20x^{3}}{3}-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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

$$3x^{4}+\frac{20x^{3}}{3}-\frac{3x^{2}}{2}+\int -5\mathrm{d}x$$

Find the integral of -5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$3x^{4}+\frac{20x^{3}}{3}-\frac{3x^{2}}{2}-5x$$

Simplify.

$$-5x-\frac{3x^{2}}{2}+\frac{20x^{3}}{3}+3x^{4}$$

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.

$$-5x-\frac{3x^{2}}{2}+\frac{20x^{3}}{3}+3x^{4}+С$$

Differentiate w.r.t. x

$$\left(3x+5\right)\left(4x^{2}-1\right)$$