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

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

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

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

\frac{1}{20} arcsec (\frac{1}{20} x) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{20} d u

= \int \frac{1}{20} d u

Take the constant out:

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

= \frac{1}{20} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{20} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{20} x)

= \frac{1}{20} \cdot 1 \cdot arcsec (\frac{1}{20} x)

Simplify

= \frac{1}{20} arcsec (\frac{1}{20} x)

Add a constant to the solution

= \frac{1}{20} arcsec (\frac{1}{20} x) + C

\int \frac{x ^{2} + 4}{x ^{3}} d x

x^{3} e^{x^{2}}

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\int \frac{2 x ^{3} + 5 x}{x ^{4} + 5 x ^{2} + 4} d x

\int x^{2} y^{3} - \frac{1}{1 + 9 x ^{2}} d x

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