What is the integral from 1 to 6 of 1/(1+sqrt(3x-2)) ?

The integral from 1 to 6 of 1/(1+sqrt(3x-2)) is (-2ln(5)+2ln(2))/3+2

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integral from 1 to 6 of 1/(1+sqrt(3x-2))

\int_{1}^{6} \frac{1}{1 + 3 x - 2} d x

\frac{- 2 ln ( 5 ) + 2 ln ( 2 )}{3} + 2

Decimal

1.38913 \dots

Solution steps

\int_{1}^{6} \frac{1}{1 + 3 x - 2} d x

Apply u-substitution

: \int_{1}^{16} \frac{1}{3 ( 1 + u )} d u

= \int_{1}^{16} \frac{1}{3 ( 1 + u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{1}^{16} \frac{1}{1 + u} d u

Apply u-substitution

: \int_{2}^{5} \frac{2 ( v - 1 )}{v} d v

= \frac{1}{3} \cdot \int_{2}^{5} \frac{2 ( v - 1 )}{v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot 2 \cdot \int_{2}^{5} \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \frac{1}{3} \cdot 2 \cdot \int_{2}^{5} 1 - \frac{1}{v} d v

Apply the Sum Rule:

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

= \frac{1}{3} \cdot 2 (\int_{2}^{5} 1 d v - \int_{2}^{5} \frac{1}{v} d v)

\int_{2}^{5} 1 d v = 3

\int_{2}^{5} \frac{1}{v} d v = ln (5) - ln (2)

= \frac{1}{3} \cdot 2 (3 - (ln (5) - ln (2)))

Simplify

\frac{1}{3} \cdot 2 (3 - (ln (5) - ln (2))) : \frac{- 2 ln ( 5 ) + 2 ln ( 2 )}{3} + 2

= \frac{- 2 ln ( 5 ) + 2 ln ( 2 )}{3} + 2

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\int_{1}^{x} ln (x) d x

What is the integral from 1 to 6 of 1/(1+sqrt(3x-2)) ?
The integral from 1 to 6 of 1/(1+sqrt(3x-2)) is (-2ln(5)+2ln(2))/3+2