What is the integral from-1 to 0 of (5-3x)/(4x-3) ?

The integral from-1 to 0 of (5-3x)/(4x-3) is (11ln(3)-11ln(7)-12)/(16)

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integral from-1 to 0 of (5-3x)/(4x-3)

\int_{- 1}^{0} \frac{5 - 3 x}{4 x - 3} d x

\frac{11 ln ( 3 ) - 11 ln ( 7 ) - 12}{16}

Decimal

- 1.33251 \dots

Solution steps

\int_{- 1}^{0} \frac{5 - 3 x}{4 x - 3} d x

Expand

\frac{5 - 3 x}{4 x - 3} : \frac{5}{4 x - 3} - \frac{3 x}{4 x - 3}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{0} \frac{5}{4 x - 3} d x - \int_{- 1}^{0} \frac{3 x}{4 x - 3} d x

\int_{- 1}^{0} \frac{5}{4 x - 3} d x = \frac{5}{4} (ln (3) - ln (7))

\int_{- 1}^{0} \frac{3 x}{4 x - 3} d x = \frac{3}{16} (4 + 3 (ln (3) - ln (7)))

= \frac{5}{4} (ln (3) - ln (7)) - \frac{3}{16} (4 + 3 (ln (3) - ln (7)))

Simplify

\frac{5}{4} (ln (3) - ln (7)) - \frac{3}{16} (4 + 3 (ln (3) - ln (7))) : \frac{11 ln ( 3 ) - 11 ln ( 7 ) - 12}{16}

= \frac{11 ln ( 3 ) - 11 ln ( 7 ) - 12}{16}

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\int_{0}^{1} x + x^{2} d x

What is the integral from-1 to 0 of (5-3x)/(4x-3) ?
The integral from-1 to 0 of (5-3x)/(4x-3) is (11ln(3)-11ln(7)-12)/(16)