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

The integral of sqrt(1+1/4 x) is 1/3 (4+x)^{3/2}+C

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integral of sqrt(1+1/4 x)

\int 1 + \frac{1}{4} x d x

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

Solution steps

\int 1 + \frac{1}{4} x d x

Simplify

1 + \frac{1}{4} x : \frac{1}{2} 4 + x

= \int \frac{1}{2} 4 + 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 4 + x d x

Apply u-substitution

: \int u d u

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 + x

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

Simplify

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

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

Add a constant to the solution

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

(2.4) (3.6)

\frac{d}{d x} (ln (8 x^{2} + 1))

\int \frac{- 3 x ^{3} + 2 x ^{2} - 3 x - 6}{x ^{3} + x} d x

n = 0 \sum \infty 3^{n} x^{2 n}

\frac{1}{3} x^{3} - \frac{16}{x} - \frac{40}{3}

What is the integral of sqrt(1+1/4 x) ?
The integral of sqrt(1+1/4 x) is 1/3 (4+x)^{3/2}+C