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

The integral of ((x^2-1))/(sqrt(2x-1)) is 1/15 (3x^2+2x-13)sqrt(2x-1)+C

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

\int \frac{( x ^{2} - 1 )}{2 x - 1} d x

\frac{1}{15} (3 x^{2} + 2 x - 13) 2 x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{( u ^{2} + 1 ) ^{2}}{4} - 1 d u

= \int \frac{( u ^{2} + 1 ) ^{2}}{4} - 1 d u

Simplify

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

= \int \frac{u ^{4} + 2 u ^{2} - 3}{4} d u

Take the constant out:

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

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

\int u^{4} d u = \frac{u ^{5}}{5}

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int 3 d u = 3 u

= \frac{1}{4} (\frac{u ^{5}}{5} + \frac{2 u ^{3}}{3} - 3 u)

Substitute back

u = 2 x - 1

= \frac{1}{4} (\frac{( 2 x - 1 ) ^{5}}{5} + \frac{2 ( 2 x - 1 ) ^{3}}{3} - 3 2 x - 1)

Simplify

\frac{1}{4} (\frac{( 2 x - 1 ) ^{5}}{5} + \frac{2 ( 2 x - 1 ) ^{3}}{3} - 3 2 x - 1) : \frac{1}{15} (3 x^{2} + 2 x - 13) 2 x - 1

= \frac{1}{15} (3 x^{2} + 2 x - 13) 2 x - 1

Add a constant to the solution

= \frac{1}{15} (3 x^{2} + 2 x - 13) 2 x - 1 + C

x \to 0 lim (\frac{0}{x ^{2} ∣ x ∣})

\frac{d y}{d t} = t^{2} + 20 t^{2} y

\int \frac{1}{9 z ^{2} - 25} d z

y = 2 t (3 t^{2} - 5)^{4}

\int e^{5 x} cos (9 x) d x

What is the integral of ((x^2-1))/(sqrt(2x-1)) ?
The integral of ((x^2-1))/(sqrt(2x-1)) is 1/15 (3x^2+2x-13)sqrt(2x-1)+C