Solve ∫ x^2/9-9a^2/frac{x{9}-a}=

$$\int \frac { \frac { x ^ { 2 } } { 9 } - 9 a ^ { 2 } } { \frac { x } { 9 } - a } =$$

Evaluate

$$9ax+\frac{x^{2}}{2}+С$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -9a^{2} times \frac{9}{9}.

$$\int \frac{\frac{x^{2}}{9}+\frac{9\left(-9\right)a^{2}}{9}}{\frac{x}{9}-a}\mathrm{d}x$$

Since \frac{x^{2}}{9} and \frac{9\left(-9\right)a^{2}}{9} have the same denominator, add them by adding their numerators.

$$\int \frac{\frac{x^{2}+9\left(-9\right)a^{2}}{9}}{\frac{x}{9}-a}\mathrm{d}x$$

Do the multiplications in x^{2}+9\left(-9\right)a^{2}.

$$\int \frac{\frac{x^{2}-81a^{2}}{9}}{\frac{x}{9}-a}\mathrm{d}x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply a times \frac{9}{9}.

$$\int \frac{\frac{x^{2}-81a^{2}}{9}}{\frac{x}{9}-\frac{9a}{9}}\mathrm{d}x$$

Since \frac{x}{9} and \frac{9a}{9} have the same denominator, subtract them by subtracting their numerators.

$$\int \frac{\frac{x^{2}-81a^{2}}{9}}{\frac{x-9a}{9}}\mathrm{d}x$$

Divide \frac{x^{2}-81a^{2}}{9} by \frac{x-9a}{9} by multiplying \frac{x^{2}-81a^{2}}{9} by the reciprocal of \frac{x-9a}{9}.

$$\int \frac{\left(x^{2}-81a^{2}\right)\times 9}{9\left(x-9a\right)}\mathrm{d}x$$

Cancel out 9 in both numerator and denominator.

$$\int \frac{x^{2}-81a^{2}}{x-9a}\mathrm{d}x$$

Factor the expressions that are not already factored in \frac{x^{2}-81a^{2}}{x-9a}.

$$\int \frac{\left(x-9a\right)\left(x+9a\right)}{x-9a}\mathrm{d}x$$

Cancel out x-9a in both numerator and denominator.

$$\int x+9a\mathrm{d}x$$

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int 9a\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x\mathrm{d}x+9\int a\mathrm{d}x$$

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}.

$$\frac{x^{2}}{2}+9\int a\mathrm{d}x$$

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

$$\frac{x^{2}}{2}+9ax$$

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.

$$\frac{x^{2}}{2}+9ax+С$$

Differentiate w.r.t. x

$$x+9a$$