What is the integral from 0 to 5 of x/((1+x)^{3/4)} ?

The integral from 0 to 5 of x/((1+x)^{3/4)} is (24*6^{1/4}-4)/5-4*6^{1/4}+4

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integral from 0 to 5 of x/((1+x)^{3/4)}

\int_{0}^{5} \frac{x}{( 1 + x ) ^{\frac{3}{4}}} d x

\frac{24 \cdot 6 ^{\frac{1}{4}} - 4}{5} - 4 \cdot 6^{\frac{1}{4}} + 4

Decimal Notation

4.45206 \dots

Solution steps

\int_{0}^{5} \frac{x}{( 1 + x ) ^{\frac{3}{4}}} d x

Apply u-substitution

: \int_{1}^{6} \frac{u - 1}{u ^{\frac{3}{4}}} d u

= \int_{1}^{6} \frac{u - 1}{u ^{\frac{3}{4}}} d u

Expand

\frac{u - 1}{u ^{\frac{3}{4}}} : u^{\frac{1}{4}} - \frac{1}{u ^{\frac{3}{4}}}

= \int_{1}^{6} u^{\frac{1}{4}} - \frac{1}{u ^{\frac{3}{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_{1}^{6} u^{\frac{1}{4}} d u - \int_{1}^{6} \frac{1}{u ^{\frac{3}{4}}} d u

\int_{1}^{6} u^{\frac{1}{4}} d u = \frac{24 \cdot 6 ^{\frac{1}{4}} - 4}{5}

\int_{1}^{6} \frac{1}{u ^{\frac{3}{4}}} d u = 4 \cdot 6^{\frac{1}{4}} - 4

= \frac{24 \cdot 6 ^{\frac{1}{4}} - 4}{5} - (4 \cdot 6^{\frac{1}{4}} - 4)

Simplify

= \frac{24 \cdot 6 ^{\frac{1}{4}} - 4}{5} - 4 \cdot 6^{\frac{1}{4}} + 4

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What is the integral from 0 to 5 of x/((1+x)^{3/4)} ?
The integral from 0 to 5 of x/((1+x)^{3/4)} is (24*6^{1/4}-4)/5-4*6^{1/4}+4