What is the integral from 0 to 1 of (1+12x)/(1+3x) ?

The integral from 0 to 1 of (1+12x)/(1+3x) is 4-2ln(2)

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integral from 0 to 1 of (1+12x)/(1+3x)

\int_{0}^{1} \frac{1 + 12 x}{1 + 3 x} d x

4 - 2 ln (2)

Decimal

2.61370 \dots

Solution steps

\int_{0}^{1} \frac{1 + 12 x}{1 + 3 x} d x

Long division

\frac{1 + 12 x}{1 + 3 x} : 4 - \frac{3}{3 x + 1}

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

Apply the Sum Rule:

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

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

\int_{0}^{1} 4 d x = 4

\int_{0}^{1} \frac{3}{3 x + 1} d x = 2 ln (2)

= 4 - 2 ln (2)

What is the integral from 0 to 1 of (1+12x)/(1+3x) ?
The integral from 0 to 1 of (1+12x)/(1+3x) is 4-2ln(2)