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

The integral of x/(sqrt(4x^2-16)) is 1/2 sqrt(x^2-4)+C

Popular >

integral of x/(sqrt(4x^2-16))

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

\frac{1}{2} x^{2} - 4 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 4 x^{2} - 16

= \frac{1}{8} \cdot 2 (4 x^{2} - 16)^{\frac{1}{2}}

Simplify

\frac{1}{8} \cdot 2 (4 x^{2} - 16)^{\frac{1}{2}} : \frac{1}{2} x^{2} - 4

= \frac{1}{2} x^{2} - 4

Add a constant to the solution

= \frac{1}{2} x^{2} - 4 + C

\int \frac{( x ^{2} + 4 ) ^{2}}{x ^{2}} d x

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

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

\int 3 e^{- 3 t} d t

\int x a r cos (x) d x

What is the integral of x/(sqrt(4x^2-16)) ?
The integral of x/(sqrt(4x^2-16)) is 1/2 sqrt(x^2-4)+C