What is the integral of x/(sqrt(1+8x)) ?

The integral of x/(sqrt(1+8x)) is 1/48 (4x-1)sqrt(8x+1)+C

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integral of x/(sqrt(1+8x))

\int \frac{x}{1 + 8 x} d x

\frac{1}{48} (4 x - 1) 8 x + 1 + C

Solution steps

\int \frac{x}{1 + 8 x} d x

Apply u-substitution

: \int \frac{u ^{2} - 1}{32} d u

= \int \frac{u ^{2} - 1}{32} d u

Take the constant out:

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

= \frac{1}{32} \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}{32} (\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}{32} (\frac{u ^{3}}{3} - u)

Substitute back

u = 1 + 8 x

= \frac{1}{32} (\frac{( 1 + 8 x ) ^{3}}{3} - 1 + 8 x)

Simplify

\frac{1}{32} (\frac{( 1 + 8 x ) ^{3}}{3} - 1 + 8 x) : \frac{1}{48} (4 x - 1) 8 x + 1

= \frac{1}{48} (4 x - 1) 8 x + 1

Add a constant to the solution

= \frac{1}{48} (4 x - 1) 8 x + 1 + C

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What is the integral of x/(sqrt(1+8x)) ?
The integral of x/(sqrt(1+8x)) is 1/48 (4x-1)sqrt(8x+1)+C