What is area y=(13)/(x^2+5x+6),y=0,x=0,x=1 ?

The answer to area y=(13)/(x^2+5x+6),y=0,x=0,x=1 is 13(-3ln(2)+2ln(3))

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area y=(13)/(x^2+5x+6),y=0,x=0,x=1

area

y = \frac{13}{x ^{2} + 5 x + 6}, y = 0, x = 0, x = 1

13 (- 3 ln (2) + 2 ln (3))

Decimal Notation

1.53117 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} \frac{13}{∣ x ^{2} + 5 x + 6 ∣} d x

Solve

\int_{0}^{1} \frac{13}{∣ x ^{2} + 5 x + 6 ∣} d x : 13 (- 3 ln (2) + 2 ln (3))

The area is:

= 13 (- 3 ln (2) + 2 ln (3))

12 y^{^{''}} - 12 y^{^{'}} + 78 y = 0

\int 5 e^{- 0.5 x} d x

\frac{d}{d x} (65 - 55 cos (\frac{π x}{15}))

\frac{\partial}{\partial x} (\frac{x ^{2} + 4}{y - 3})

y^{^{'}} = \frac{y ^{2} - y}{x}

What is area y=(13)/(x^2+5x+6),y=0,x=0,x=1 ?
The answer to area y=(13)/(x^2+5x+6),y=0,x=0,x=1 is 13(-3ln(2)+2ln(3))