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

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

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

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

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

Solution steps

\int \frac{x}{4 x + 1} 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 = 4 x + 1

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

Simplify

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

(6 x y - 2 y^{2} + 1) d x + (3 x^{2} - 4 x y) d y = 0

\int \frac{5}{6 x + 5} d x

\frac{x}{7 x ^{2} + 1}

f (x) = e^{2 e^{x}}

\frac{d}{d x} (e^{x} (a cos (x) + b sin (x)))

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