What is the integral of (x+4)(x+3)^{1/4} ?

The integral of (x+4)(x+3)^{1/4} is 4/9 (x+3)^{9/4}+4/5 (x+3)^{5/4}+C

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integral of (x+4)(x+3)^{1/4}

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

\frac{4}{9} (x + 3)^{\frac{9}{4}} + \frac{4}{5} (x + 3)^{\frac{5}{4}} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}}

= \frac{4}{9} v^{\frac{9}{4}} + \frac{4}{5} v^{\frac{5}{4}}

Substitute back

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

Simplify

\frac{4}{9} (x + 4 - 1)^{\frac{9}{4}} + \frac{4}{5} (x + 4 - 1)^{\frac{5}{4}} : \frac{4}{9} (x + 3)^{\frac{9}{4}} + \frac{4}{5} (x + 3)^{\frac{5}{4}}

= \frac{4}{9} (x + 3)^{\frac{9}{4}} + \frac{4}{5} (x + 3)^{\frac{5}{4}}

Add a constant to the solution

= \frac{4}{9} (x + 3)^{\frac{9}{4}} + \frac{4}{5} (x + 3)^{\frac{5}{4}} + C

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

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What is the integral of (x+4)(x+3)^{1/4} ?
The integral of (x+4)(x+3)^{1/4} is 4/9 (x+3)^{9/4}+4/5 (x+3)^{5/4}+C