What is the integral of 1/(xsqrt(400x^2-400)) ?

The integral of 1/(xsqrt(400x^2-400)) is 1/20 arctan(sqrt(x^2-1))+C

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integral of 1/(xsqrt(400x^2-400))

\int \frac{1}{x 400 x ^{2} - 400} d x

\frac{1}{20} arctan (x^{2} - 1) + C

Solution steps

\int \frac{1}{x 400 x ^{2} - 400} d x

Apply Integration By Parts

: \frac{1}{20 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{20 x ( x ^{2} - 1 ) x ^{2} - 1} d x

= \frac{1}{20 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{20 x ( x ^{2} - 1 ) x ^{2} - 1} d x

\int - \frac{2 x ^{2} - 1}{20 x ( x ^{2} - 1 ) x ^{2} - 1} d x = - \frac{1}{20} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))

= \frac{1}{20 x ^{2} - 1} - (- \frac{1}{20} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1)))

Simplify

\frac{1}{20 x ^{2} - 1} - (- \frac{1}{20} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))) : \frac{1}{20} arctan (x^{2} - 1)

= \frac{1}{20} arctan (x^{2} - 1)

Add a constant to the solution

= \frac{1}{20} arctan (x^{2} - 1) + C

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\frac{d}{d x} (60 x^{4} - 180 x^{3} + 180 x^{2} - 60 x)

What is the integral of 1/(xsqrt(400x^2-400)) ?
The integral of 1/(xsqrt(400x^2-400)) is 1/20 arctan(sqrt(x^2-1))+C