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

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

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

\int \frac{1}{x x ^{2} - π} d x

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

Solution steps

\int \frac{1}{x x ^{2} - π} d x

Apply Trigonometric Substitution

: \int \frac{1}{π} d u

= \int \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{π} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{π} x)

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

Simplify

= \frac{1}{π} arcsec (\frac{1}{π} x)

Add a constant to the solution

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

\int_{0}^{1} e^{x} - x d x

\int x sin (3 x) d x

y = 2 cos (x), y = 6 - 4 cos (x), 0 \leq x \leq 2 π

f (x) = 18 x^{2} - 14 x + 9

\frac{d}{d x} (ln (x^{2} + 2 x + 1))

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