Solve ∫ [a^2-5a+6/a^2+7a+6:(2a+10/a+1-a-1)+1/a+3]*2a^2+5a-3/2a^2

$$\int [ \frac { a ^ { 2 } - 5 a + 6 } { a ^ { 2 } + 7 a + 6 } : ( \frac { 2 a + 10 } { a + 1 } - a - 1 ) + \frac { 1 } { a + 3 } ] \cdot \frac { 2 a ^ { 2 } + 5 a - 3 } { 2 a ^ { 2 } }$$

Evaluate

$$\frac{8ax-4x}{\left(a+6\right)a^{2}}+С$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -a-1 times \frac{a+1}{a+1}.

$$\int \left(\frac{\frac{a^{2}-5a+6}{a^{2}+7a+6}}{\frac{2a+10}{a+1}+\frac{\left(-a-1\right)\left(a+1\right)}{a+1}}+\frac{1}{a+3}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Since \frac{2a+10}{a+1} and \frac{\left(-a-1\right)\left(a+1\right)}{a+1} have the same denominator, add them by adding their numerators.

$$\int \left(\frac{\frac{a^{2}-5a+6}{a^{2}+7a+6}}{\frac{2a+10+\left(-a-1\right)\left(a+1\right)}{a+1}}+\frac{1}{a+3}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Do the multiplications in 2a+10+\left(-a-1\right)\left(a+1\right).

$$\int \left(\frac{\frac{a^{2}-5a+6}{a^{2}+7a+6}}{\frac{2a+10-a^{2}-a-a-1}{a+1}}+\frac{1}{a+3}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Combine like terms in 2a+10-a^{2}-a-a-1.

$$\int \left(\frac{\frac{a^{2}-5a+6}{a^{2}+7a+6}}{\frac{9-a^{2}}{a+1}}+\frac{1}{a+3}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Divide \frac{a^{2}-5a+6}{a^{2}+7a+6} by \frac{9-a^{2}}{a+1} by multiplying \frac{a^{2}-5a+6}{a^{2}+7a+6} by the reciprocal of \frac{9-a^{2}}{a+1}.

$$\int \left(\frac{\left(a^{2}-5a+6\right)\left(a+1\right)}{\left(a^{2}+7a+6\right)\left(9-a^{2}\right)}+\frac{1}{a+3}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Factor the expressions that are not already factored in \frac{\left(a^{2}-5a+6\right)\left(a+1\right)}{\left(a^{2}+7a+6\right)\left(9-a^{2}\right)}.

$$\int \left(\frac{\left(a-3\right)\left(a-2\right)\left(a+1\right)}{\left(a-3\right)\left(-a-3\right)\left(a+1\right)\left(a+6\right)}+\frac{1}{a+3}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Cancel out \left(a-3\right)\left(a+1\right) in both numerator and denominator.

$$\int \left(\frac{a-2}{\left(-a-3\right)\left(a+6\right)}+\frac{1}{a+3}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(-a-3\right)\left(a+6\right) and a+3 is \left(a+3\right)\left(a+6\right). Multiply \frac{a-2}{\left(-a-3\right)\left(a+6\right)} times \frac{-1}{-1}. Multiply \frac{1}{a+3} times \frac{a+6}{a+6}.

$$\int \left(\frac{-\left(a-2\right)}{\left(a+3\right)\left(a+6\right)}+\frac{a+6}{\left(a+3\right)\left(a+6\right)}\right)\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Since \frac{-\left(a-2\right)}{\left(a+3\right)\left(a+6\right)} and \frac{a+6}{\left(a+3\right)\left(a+6\right)} have the same denominator, add them by adding their numerators.

$$\int \frac{-\left(a-2\right)+a+6}{\left(a+3\right)\left(a+6\right)}\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Do the multiplications in -\left(a-2\right)+a+6.

$$\int \frac{-a+2+a+6}{\left(a+3\right)\left(a+6\right)}\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Combine like terms in -a+2+a+6.

$$\int \frac{8}{\left(a+3\right)\left(a+6\right)}\times \frac{2a^{2}+5a-3}{2a^{2}}\mathrm{d}x$$

Multiply \frac{8}{\left(a+3\right)\left(a+6\right)} times \frac{2a^{2}+5a-3}{2a^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\int \frac{8\left(2a^{2}+5a-3\right)}{\left(a+3\right)\left(a+6\right)\times 2a^{2}}\mathrm{d}x$$

Cancel out 2 in both numerator and denominator.

$$\int \frac{4\left(2a^{2}+5a-3\right)}{\left(a+3\right)\left(a+6\right)a^{2}}\mathrm{d}x$$

Factor the expressions that are not already factored in \frac{4\left(2a^{2}+5a-3\right)}{\left(a+3\right)\left(a+6\right)a^{2}}.

$$\int \frac{4\left(2a-1\right)\left(a+3\right)}{\left(a+3\right)\left(a+6\right)a^{2}}\mathrm{d}x$$

Cancel out a+3 in both numerator and denominator.

$$\int \frac{4\left(2a-1\right)}{\left(a+6\right)a^{2}}\mathrm{d}x$$

Use the distributive property to multiply 4 by 2a-1.

$$\int \frac{8a-4}{\left(a+6\right)a^{2}}\mathrm{d}x$$

Use the distributive property to multiply a+6 by a^{2}.

$$\int \frac{8a-4}{a^{3}+6a^{2}}\mathrm{d}x$$

Find the integral of \frac{8a-4}{a^{3}+6a^{2}} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{8a-4}{a^{3}+6a^{2}}x$$

Simplify.

$$\frac{\left(8a-4\right)x}{a^{3}+6a^{2}}$$

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.

$$\frac{\left(8a-4\right)x}{a^{3}+6a^{2}}+С$$

Differentiate w.r.t. x

$$\frac{4\left(2a-1\right)}{\left(a+6\right)a^{2}}$$