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

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

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

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

\frac{1}{2} 2 x^{2} - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} 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 \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 x^{2} - 1

= \frac{1}{4} \cdot 2 (2 x^{2} - 1)^{\frac{1}{2}}

Simplify

\frac{1}{4} \cdot 2 (2 x^{2} - 1)^{\frac{1}{2}} : \frac{1}{2} 2 x^{2} - 1

= \frac{1}{2} 2 x^{2} - 1

Add a constant to the solution

= \frac{1}{2} 2 x^{2} - 1 + C

\frac{d y}{d x} - (\frac{4}{x}) y = x^{5} e^{x}

x \to 2 lim (\frac{x ^{2} - 2}{\frac{5 x}{7} - \frac{5 2}{7}})

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 5)^{5})

\frac{d}{d x} (\frac{x - 5}{x ^{2}})

x \to 6 - lim (c x^{2} + 3 x)

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