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

The integral of 1/(xsqrt(196x^2-196)) is 1/14 arctan(sqrt(x^2-1))+C

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

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

\frac{1}{14} arctan (x^{2} - 1) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{14 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{14 x ( x ^{2} - 1 ) x ^{2} - 1} d x

= \frac{1}{14 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{14 x ( x ^{2} - 1 ) x ^{2} - 1} d x

\int - \frac{2 x ^{2} - 1}{14 x ( x ^{2} - 1 ) x ^{2} - 1} d x = - \frac{1}{14} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))

= \frac{1}{14 x ^{2} - 1} - (- \frac{1}{14} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1)))

Simplify

\frac{1}{14 x ^{2} - 1} - (- \frac{1}{14} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))) : \frac{1}{14} arctan (x^{2} - 1)

= \frac{1}{14} arctan (x^{2} - 1)

Add a constant to the solution

= \frac{1}{14} arctan (x^{2} - 1) + C

\frac{x ^{2}}{5 + 4 x}

e^{x}, e^{- 3 x}, ln (4)

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

\frac{d}{d x} (\frac{3 ( x ^{2} + 2 ) ^{3}}{5})

\frac{d x}{d t} = a x

What is the integral of 1/(xsqrt(196x^2-196)) ?
The integral of 1/(xsqrt(196x^2-196)) is 1/14 arctan(sqrt(x^2-1))+C