What is the integral of x/(sqrt(196x^2-1)) ?

The integral of x/(sqrt(196x^2-1)) is 1/196 sqrt(196x^2-1)+C

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integral of x/(sqrt(196x^2-1))

\int \frac{x}{196 x ^{2} - 1} d x

\frac{1}{196} 196 x^{2} - 1 + C

Solution steps

\int \frac{x}{196 x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{392 u} d u

= \int \frac{1}{392 u} d u

Take the constant out:

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

= \frac{1}{392} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{392} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 196 x^{2} - 1

= \frac{1}{392} \cdot 2 (196 x^{2} - 1)^{\frac{1}{2}}

Simplify

\frac{1}{392} \cdot 2 (196 x^{2} - 1)^{\frac{1}{2}} : \frac{1}{196} 196 x^{2} - 1

= \frac{1}{196} 196 x^{2} - 1

Add a constant to the solution

= \frac{1}{196} 196 x^{2} - 1 + C

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What is the integral of x/(sqrt(196x^2-1)) ?
The integral of x/(sqrt(196x^2-1)) is 1/196 sqrt(196x^2-1)+C