What is the integral from 0 to 1 of (1+6t)/(1+2t) ?

The integral from 0 to 1 of (1+6t)/(1+2t) is -ln(3)+3

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integral from 0 to 1 of (1+6t)/(1+2t)

\int_{0}^{1} \frac{1 + 6 t}{1 + 2 t} d t

- ln (3) + 3

Decimal

1.90138 \dots

Solution steps

\int_{0}^{1} \frac{1 + 6 t}{1 + 2 t} d t

Expand

\frac{1 + 6 t}{1 + 2 t} : \frac{1}{1 + 2 t} + \frac{6 t}{1 + 2 t}

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} \frac{1}{1 + 2 t} d t + \int_{0}^{1} \frac{6 t}{1 + 2 t} d t

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

\int_{0}^{1} \frac{6 t}{1 + 2 t} d t = 3 - \frac{3}{2} ln (3)

= \frac{1}{2} ln (3) + 3 - \frac{3}{2} ln (3)

Simplify

\frac{1}{2} ln (3) + 3 - \frac{3}{2} ln (3) : - ln (3) + 3

= - ln (3) + 3

What is the integral from 0 to 1 of (1+6t)/(1+2t) ?
The integral from 0 to 1 of (1+6t)/(1+2t) is -ln(3)+3