What is the integral of 1/(xsqrt(x^2-49)) ?

The integral of 1/(xsqrt(x^2-49)) is 1/7 arcsec(1/7 x)+C

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integral of 1/(xsqrt(x^2-49))

\int \frac{1}{x x ^{2} - 49} d x

\frac{1}{7} arcsec (\frac{1}{7} x) + C

Solution steps

\int \frac{1}{x x ^{2} - 49} d x

Apply Trigonometric Substitution

: \int \frac{1}{7} d u

= \int \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{7} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{7} x)

= \frac{1}{7} \cdot 1 \cdot arcsec (\frac{1}{7} x)

Simplify

= \frac{1}{7} arcsec (\frac{1}{7} x)

Add a constant to the solution

= \frac{1}{7} arcsec (\frac{1}{7} x) + C

x \to + 0 lim (\frac{e ^{- 1 x} - 1}{x})

\frac{d}{d x} y = x^{3} (2 - x^{2})

y = x^{0.95}

a^{4} + cos^{4} (x)

What is the integral of 1/(xsqrt(x^2-49)) ?
The integral of 1/(xsqrt(x^2-49)) is 1/7 arcsec(1/7 x)+C