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

The integral of (sqrt(2x-1))/x is 2(-arctan(sqrt(2x-1))+sqrt(2x-1))+C

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integral of (sqrt(2x-1))/x

\int \frac{2 x - 1}{x} d x

2 (- arctan (2 x - 1) + 2 x - 1) + C

Solution steps

\int \frac{2 x - 1}{x} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} + 1} d u

= \int \frac{2 u ^{2}}{u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= 2 \cdot \int - \frac{1}{u ^{2} + 1} + 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

= 2 (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= 2 (- arctan (u) + u)

Substitute back

u = 2 x - 1

= 2 (- arctan (2 x - 1) + 2 x - 1)

Add a constant to the solution

= 2 (- arctan (2 x - 1) + 2 x - 1) + C

x^{2} sin (4 x)

\frac{\partial}{\partial x} (6 x^{5} y^{6} + 8 x^{7} y^{8})

\frac{\partial}{\partial x} (x arcsin (3 x))

f (x) = sin (4 x^{2})

x \to 0 lim (\frac{cos ( 2 x )}{sin ( 2 x )})

What is the integral of (sqrt(2x-1))/x ?
The integral of (sqrt(2x-1))/x is 2(-arctan(sqrt(2x-1))+sqrt(2x-1))+C