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

The integral from 1 to 6 of 1/(2sqrt(x+3)+x) is (-3ln(5)+ln(3)+2ln(12))/2

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

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

\frac{- 3 ln ( 5 ) + ln ( 3 ) + 2 ln ( 12 )}{2}

Decimal Notation

0.62005 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} \frac{2 u}{2 u + u ^{2} - 3} d u

= \int_{2}^{3} \frac{2 u}{2 u + u ^{2} - 3} d u

Take the constant out:

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

= 2 \cdot \int_{2}^{3} \frac{u}{2 u + u ^{2} - 3} d u

Complete the square

2 u + u^{2} - 3 : (u + 1)^{2} - 4

= 2 \cdot \int_{2}^{3} \frac{u}{( u + 1 ) ^{2} - 4} d u

Apply u-substitution

: \int_{3}^{4} \frac{v - 1}{v ^{2} - 4} d v

= 2 \cdot \int_{3}^{4} \frac{v - 1}{v ^{2} - 4} d v

Expand

\frac{v - 1}{v ^{2} - 4} : \frac{v}{v ^{2} - 4} - \frac{1}{v ^{2} - 4}

Apply the Sum Rule:

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

= 2 (\int_{3}^{4} \frac{v}{v ^{2} - 4} d v - \int_{3}^{4} \frac{1}{v ^{2} - 4} d v)

\int_{3}^{4} \frac{v}{v ^{2} - 4} d v = \frac{1}{2} (ln (12) - ln (5))

\int_{3}^{4} \frac{1}{v ^{2} - 4} d v = - \frac{ln ( 3 ) - ln ( 5 )}{4}

= 2 (\frac{1}{2} (ln (12) - ln (5)) - (- \frac{ln ( 3 ) - ln ( 5 )}{4}))

Simplify

2 (\frac{1}{2} (ln (12) - ln (5)) - (- \frac{ln ( 3 ) - ln ( 5 )}{4})) : \frac{- 3 ln ( 5 ) + ln ( 3 ) + 2 ln ( 12 )}{2}

= \frac{- 3 ln ( 5 ) + ln ( 3 ) + 2 ln ( 12 )}{2}

\frac{d}{d x} (7 x (x^{3} - 8 x + 2))

x \frac{d y}{d x} - 2 y = 5 x^{3}

\frac{\partial}{\partial x} (\frac{x ^{2}}{sin ^{2} ( x )})

\frac{\partial}{\partial z} (\frac{( x + y )}{y + z})

x \to 2 + lim (- \frac{5}{x - 2})

What is the integral from 1 to 6 of 1/(2sqrt(x+3)+x) ?
The integral from 1 to 6 of 1/(2sqrt(x+3)+x) is (-3ln(5)+ln(3)+2ln(12))/2