What is the integral of ((sqrt(x)+1)^4)/(2sqrt(x)) ?

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

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integral of ((sqrt(x)+1)^4)/(2sqrt(x))

\int \frac{( x + 1 ) ^{4}}{2 x} d x

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

Solution steps

\int \frac{( x + 1 ) ^{4}}{2 x} d x

Take the constant out:

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

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

Apply u-substitution

: \int 2 u^{4} d u

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{2} \cdot 2 \cdot \frac{u ^{5}}{5}

Substitute back

u = x + 1

= \frac{1}{2} \cdot 2 \cdot \frac{( x + 1 ) ^{5}}{5}

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{( x + 1 ) ^{5}}{5} : \frac{1}{5} (x + 1)^{5}

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

Add a constant to the solution

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

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\int_{1}^{16} \frac{1}{3 - x} d x

4 t \cdot \frac{d y}{d t} - y = t

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