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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int (u - 1) u^{\frac{1}{4}} d u

Expand

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

= \int u^{\frac{5}{4}} - u^{\frac{1}{4}} 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

= \int u^{\frac{5}{4}} d u - \int u^{\frac{1}{4}} d u

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

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

= \frac{4}{9} u^{\frac{9}{4}} - \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 1 + x

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{cos ( x )}{sin ( x ) + cos ( x )})

f (x) = \frac{e ^{x}}{x ^{2} + 4}, x = 1

f (x) = \frac{x + 1}{x - 2}

\int 0 d x

x^{4} - 3 x^{3} + 2 x^{2} - x + 1

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