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

The integral of x/(sqrt(1-4x)) is -1/12 (2x+1)sqrt(-4x+1)+C

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

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

- \frac{1}{12} (2 x + 1) - 4 x + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} - 1}{8} d u

= \int \frac{u ^{2} - 1}{8} 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 u^{2} - 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

= \frac{1}{8} (\int u^{2} d u - \int 1 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{1}{8} (\frac{u ^{3}}{3} - u)

Substitute back

u = 1 - 4 x

= \frac{1}{8} (\frac{( 1 - 4 x ) ^{3}}{3} - 1 - 4 x)

Simplify

\frac{1}{8} (\frac{( 1 - 4 x ) ^{3}}{3} - 1 - 4 x) : - \frac{1}{12} (2 x + 1) - 4 x + 1

= - \frac{1}{12} (2 x + 1) - 4 x + 1

Add a constant to the solution

= - \frac{1}{12} (2 x + 1) - 4 x + 1 + C

\int - \frac{sin ( 3 x )}{cos ^{3} ( 3 x )} d x

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

\int \frac{x ^{2} - 5}{( 1 + x ) ^{3}} d x

\frac{1}{4 x}

\int 6 x + 12 x^{2} d x

What is the integral of x/(sqrt(1-4x)) ?
The integral of x/(sqrt(1-4x)) is -1/12 (2x+1)sqrt(-4x+1)+C