What is the integral from 1 to 5 of x/(sqrt(2x-1)) ?

The integral from 1 to 5 of x/(sqrt(2x-1)) is 16/3

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integral from 1 to 5 of x/(sqrt(2x-1))

\int_{1}^{5} \frac{x}{2 x - 1} d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{1}^{5} \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{16}{3}

= \frac{16}{3}

x \to 0 lim (cos (x) + sin (x))

(x + 2)^{2} + (y - 3)^{2} = 37, (4, 2)

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{x ^{2} - 1}, y (7) = 7

\int e^{\frac{x}{5}} d x

y^{^{'}} = 8 x^{3} + 2 \frac{y}{x}

What is the integral from 1 to 5 of x/(sqrt(2x-1)) ?
The integral from 1 to 5 of x/(sqrt(2x-1)) is 16/3