What is the integral from 0 to 5 of x/(sqrt(1+4x)) ?

The integral from 0 to 5 of x/(sqrt(1+4x)) is (9sqrt(21)+1)/(12)

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integral from 0 to 5 of x/(sqrt(1+4x))

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

\frac{9 21 + 1}{12}

Decimal Notation

3.52026 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{21} \frac{u ^{2} - 1}{8} d u

= \int_{1}^{21} \frac{u ^{2} - 1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{21} u^{2} - 1 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

= \frac{1}{8} (\int_{1}^{21} u^{2} d u - \int_{1}^{21} 1 d u)

\int_{1}^{21} u^{2} d u = 721 - \frac{1}{3}

\int_{1}^{21} 1 d u = 21 - 1

= \frac{1}{8} (721 - \frac{1}{3} - (21 - 1))

Simplify

\frac{1}{8} (721 - \frac{1}{3} - (21 - 1)) : \frac{9 21 + 1}{12}

= \frac{9 21 + 1}{12}

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

\int_{0}^{2} x - 1 d x

x^{^{''}} + 2 x = 0, x (0) = 1, x (1) = 0

13440 (2 x + 5)^{3}

y = x^{2} - 2 x - 2 P (y, x), (2, - 2)

What is the integral from 0 to 5 of x/(sqrt(1+4x)) ?
The integral from 0 to 5 of x/(sqrt(1+4x)) is (9sqrt(21)+1)/(12)