What is the integral of (11)/(2xsqrt(4x^2-16)) ?

The integral of (11)/(2xsqrt(4x^2-16)) is 11/8 arcsec(x/2)+C

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integral of (11)/(2xsqrt(4x^2-16))

\int \frac{11}{2 x 4 x ^{2} - 16} d x

\frac{11}{8} arcsec (\frac{x}{2}) + C

Solution steps

\int \frac{11}{2 x 4 x ^{2} - 16} d x

Take the constant out:

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

= \frac{11}{2} \cdot \int \frac{1}{x 4 x ^{2} - 16} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

= \frac{11}{2} \cdot \int \frac{1}{4} d u

Take the constant out:

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

= \frac{11}{2} \cdot \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{11}{2} \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{2}{4} x)

= \frac{11}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{2}{4} x)

Simplify

\frac{11}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{2}{4} x) : \frac{11}{8} arcsec (\frac{x}{2})

= \frac{11}{8} arcsec (\frac{x}{2})

Add a constant to the solution

= \frac{11}{8} arcsec (\frac{x}{2}) + C

\int \frac{( a x + b )}{( c x + d )} d x

\frac{7}{2 x ^{2}}

x \to \infty lim (arctan (x^{6} - x^{9}))

y = arctan (3 x)

f (x) = \frac{3 + 4 x ^{\frac{1}{2}}}{2 + x ^{\frac{3}{2}}}

What is the integral of (11)/(2xsqrt(4x^2-16)) ?
The integral of (11)/(2xsqrt(4x^2-16)) is 11/8 arcsec(x/2)+C