Solve ∫ (-6/10-5/6)*(-7+1)*(3/5-5)+2

$$\int ( \frac { - 6 } { 10 } - \frac { 5 } { 6 } ) \cdot ( - 7 + 1 ) \cdot ( \frac { 3 } { 5 } - 5 ) + 2$$

Evaluate

$$-\frac{896x}{25}+С$$

Solution Steps

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

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

Least common multiple of 5 and 6 is 30. Convert -\frac{3}{5} and \frac{5}{6} to fractions with denominator 30.

$$\int \left(-\frac{18}{30}-\frac{25}{30}\right)\left(-7+1\right)\left(\frac{3}{5}-5\right)+2\mathrm{d}x$$

Since -\frac{18}{30} and \frac{25}{30} have the same denominator, subtract them by subtracting their numerators.

$$\int \frac{-18-25}{30}\left(-7+1\right)\left(\frac{3}{5}-5\right)+2\mathrm{d}x$$

Subtract 25 from -18 to get -43.

$$\int -\frac{43}{30}\left(-7+1\right)\left(\frac{3}{5}-5\right)+2\mathrm{d}x$$

Add -7 and 1 to get -6.

$$\int -\frac{43}{30}\left(-6\right)\left(\frac{3}{5}-5\right)+2\mathrm{d}x$$

Express -\frac{43}{30}\left(-6\right) as a single fraction.

$$\int \frac{-43\left(-6\right)}{30}\left(\frac{3}{5}-5\right)+2\mathrm{d}x$$

Multiply -43 and -6 to get 258.

$$\int \frac{258}{30}\left(\frac{3}{5}-5\right)+2\mathrm{d}x$$

Reduce the fraction \frac{258}{30} to lowest terms by extracting and canceling out 6.

$$\int \frac{43}{5}\left(\frac{3}{5}-5\right)+2\mathrm{d}x$$

Convert 5 to fraction \frac{25}{5}.

$$\int \frac{43}{5}\left(\frac{3}{5}-\frac{25}{5}\right)+2\mathrm{d}x$$

Since \frac{3}{5} and \frac{25}{5} have the same denominator, subtract them by subtracting their numerators.

$$\int \frac{43}{5}\times \frac{3-25}{5}+2\mathrm{d}x$$

Subtract 25 from 3 to get -22.

$$\int \frac{43}{5}\left(-\frac{22}{5}\right)+2\mathrm{d}x$$

Multiply \frac{43}{5} times -\frac{22}{5} by multiplying numerator times numerator and denominator times denominator.

$$\int \frac{43\left(-22\right)}{5\times 5}+2\mathrm{d}x$$

Do the multiplications in the fraction \frac{43\left(-22\right)}{5\times 5}.

$$\int \frac{-946}{25}+2\mathrm{d}x$$

Fraction \frac{-946}{25} can be rewritten as -\frac{946}{25} by extracting the negative sign.

$$\int -\frac{946}{25}+2\mathrm{d}x$$

Convert 2 to fraction \frac{50}{25}.

$$\int -\frac{946}{25}+\frac{50}{25}\mathrm{d}x$$

Since -\frac{946}{25} and \frac{50}{25} have the same denominator, add them by adding their numerators.

$$\int \frac{-946+50}{25}\mathrm{d}x$$

Add -946 and 50 to get -896.

$$\int -\frac{896}{25}\mathrm{d}x$$

Find the integral of -\frac{896}{25} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{896x}{25}$$

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.

$$-\frac{896x}{25}+С$$

Differentiate w.r.t. x

$$-\frac{896}{25} = -35\frac{21}{25} = -35.84$$