What is the integral of x/(sqrt(6x+1)) ?

The integral of x/(sqrt(6x+1)) is 1/27 (3x-1)sqrt(6x+1)+C

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integral of x/(sqrt(6x+1))

\int \frac{x}{6 x + 1} d x

\frac{1}{27} (3 x - 1) 6 x + 1 + C

Solution steps

\int \frac{x}{6 x + 1} d x

Apply u-substitution

: \int \frac{u ^{2} - 1}{18} d u

= \int \frac{u ^{2} - 1}{18} d u

Take the constant out:

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

= \frac{1}{18} \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}{18} (\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}{18} (\frac{u ^{3}}{3} - u)

Substitute back

u = 6 x + 1

= \frac{1}{18} (\frac{( 6 x + 1 ) ^{3}}{3} - 6 x + 1)

Simplify

\frac{1}{18} (\frac{( 6 x + 1 ) ^{3}}{3} - 6 x + 1) : \frac{1}{27} (3 x - 1) 6 x + 1

= \frac{1}{27} (3 x - 1) 6 x + 1

Add a constant to the solution

= \frac{1}{27} (3 x - 1) 6 x + 1 + C

arctan (\frac{2 x}{5})

4 x 3 x^{2} + 2

\frac{x ^{5}}{10} + \frac{1}{6 x ^{3}}

t \cdot \frac{d y}{d t} = t^{3} + 14 t^{3} y

\int \frac{cos ( t )}{sin ^{4} ( t ) - 16} d t

What is the integral of x/(sqrt(6x+1)) ?
The integral of x/(sqrt(6x+1)) is 1/27 (3x-1)sqrt(6x+1)+C