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

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

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integral of (sqrt(1-4x^2))/x

\int \frac{1 - 4 x ^{2}}{x} d x

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

Solution steps

\int \frac{1 - 4 x ^{2}}{x} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} - 1} d u

= \int \frac{u ^{2}}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= - \int \frac{u ^{2}}{- u ^{2} + 1} d u

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

= - \int \frac{1}{- u ^{2} + 1} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int \frac{1}{- u ^{2} + 1} d u - \int 1 d u)

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u)

Substitute back

u = 1 - 4 x^{2}

= - (\frac{ln 1 - 4 x ^{2} + 1}{2} - \frac{ln 1 - 4 x ^{2} - 1}{2} - 1 - 4 x^{2})

Simplify

- (\frac{ln 1 - 4 x ^{2} + 1}{2} - \frac{ln 1 - 4 x ^{2} - 1}{2} - 1 - 4 x^{2}) : - \frac{1}{2} ln 1 - 4 x^{2} + 1 + \frac{1}{2} ln 1 - 4 x^{2} - 1 + 1 - 4 x^{2}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{2 x ^{2} + 2 x + 6}{x})

\frac{2}{3 t ^{2}}

\int \frac{1}{2} x^{4} d x

\frac{440 t}{( t ^{2} + 6 ) ^{3}}

f (x) = sin^{5} (3 x)

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