What is the integral from 0 to 4 of x/(sqrt(2x+1)) ?

The integral from 0 to 4 of x/(sqrt(2x+1)) is 10/3

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integral from 0 to 4 of x/(sqrt(2x+1))

\int_{0}^{4} \frac{x}{2 x + 1} d x

\frac{10}{3}

Decimal Notation

3.33333 \dots

Solution steps

\int_{0}^{4} \frac{x}{2 x + 1} d x

Apply u-substitution

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

= \int_{1}^{3} \frac{u ^{2} - 1}{2} d u

Take the constant out:

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

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

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}{2} (\int_{1}^{3} u^{2} d u - \int_{1}^{3} 1 d u)

\int_{1}^{3} u^{2} d u = \frac{26}{3}

\int_{1}^{3} 1 d u = 2

= \frac{1}{2} (\frac{26}{3} - 2)

Simplify

\frac{1}{2} (\frac{26}{3} - 2) : \frac{10}{3}

= \frac{10}{3}

x \to 0 + lim (\frac{5}{x + x ^{2}})

\frac{f ( a + h ) - f ( a - h )}{2 h}

x \to - 1 lim (∣ x - 1 ∣)

(9, - 1), (4, - 11)

y = 6, y = x^{2} - 3

What is the integral from 0 to 4 of x/(sqrt(2x+1)) ?
The integral from 0 to 4 of x/(sqrt(2x+1)) is 10/3