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

The integral of (sqrt(x)+1)^{1/4} is 2(4/9 (sqrt(x)+1)^{9/4}-4/5 (sqrt(x)+1)^{5/4})+C

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integral of (sqrt(x)+1)^{1/4}

\int (x + 1)^{\frac{1}{4}} d x

2 (\frac{4}{9} (x + 1)^{\frac{9}{4}} - \frac{4}{5} (x + 1)^{\frac{5}{4}}) + C

Solution steps

\int (x + 1)^{\frac{1}{4}} d x

Apply u-substitution

: \int (u + 1)^{\frac{1}{4}} \cdot 2 u d u

= \int (u + 1)^{\frac{1}{4}} \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int (u + 1)^{\frac{1}{4}} u d u

Apply u-substitution

: \int (v - 1) v^{\frac{1}{4}} d v

= 2 \cdot \int (v - 1) v^{\frac{1}{4}} d v

Expand

(v - 1) v^{\frac{1}{4}} : v^{\frac{5}{4}} - v^{\frac{1}{4}}

= 2 \cdot \int v^{\frac{5}{4}} - v^{\frac{1}{4}} d v

Apply the Sum Rule:

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

= 2 (\int v^{\frac{5}{4}} d v - \int v^{\frac{1}{4}} d v)

\int v^{\frac{5}{4}} d v = \frac{4}{9} v^{\frac{9}{4}}

\int v^{\frac{1}{4}} d v = \frac{4}{5} v^{\frac{5}{4}}

= 2 (\frac{4}{9} v^{\frac{9}{4}} - \frac{4}{5} v^{\frac{5}{4}})

Substitute back

= 2 (\frac{4}{9} (x + 1)^{\frac{9}{4}} - \frac{4}{5} (x + 1)^{\frac{5}{4}})

Add a constant to the solution

= 2 (\frac{4}{9} (x + 1)^{\frac{9}{4}} - \frac{4}{5} (x + 1)^{\frac{5}{4}}) + C

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What is the integral of (sqrt(x)+1)^{1/4} ?
The integral of (sqrt(x)+1)^{1/4} is 2(4/9 (sqrt(x)+1)^{9/4}-4/5 (sqrt(x)+1)^{5/4})+C