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

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

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

integral

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x}{4 ( 1 + x ) ^{\frac{3}{4}}} d x = \frac{1}{5} (x - 4) (x + 1)^{\frac{1}{4}}

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

Simplify

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

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

Add a constant to the solution

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

e^{(8 x)^{2}}

ln (1 - x)

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

(x^{2} + 5) \cdot 2^{x}

2 e^{- 2} x

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