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

The integral of xsqrt(4x+1) is 1/60 (6x-1)(4x+1)^{3/2}+C

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integral of xsqrt(4x+1)

\int x 4 x + 1 d x

\frac{1}{60} (6 x - 1) (4 x + 1)^{\frac{3}{2}} + C

Solution steps

\int x 4 x + 1 d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{16} d u

= \int \frac{u ( u - 1 )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{16} \cdot \int u^{\frac{3}{2}} - u 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}{16} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{16} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 4 x + 1

= \frac{1}{16} (\frac{2}{5} (4 x + 1)^{\frac{5}{2}} - \frac{2}{3} (4 x + 1)^{\frac{3}{2}})

Simplify

\frac{1}{16} (\frac{2}{5} (4 x + 1)^{\frac{5}{2}} - \frac{2}{3} (4 x + 1)^{\frac{3}{2}}) : \frac{1}{60} (6 x - 1) (4 x + 1)^{\frac{3}{2}}

= \frac{1}{60} (6 x - 1) (4 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{60} (6 x - 1) (4 x + 1)^{\frac{3}{2}} + C

f (x) = 5 x + 7 x

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 4

\int \frac{x ^{3} + 1}{x ^{3} + x ^{2}} d x

y = e^{- x} arctan (5 e^{x})

x^{2} ln (x^{2} + 1)

What is the integral of xsqrt(4x+1) ?
The integral of xsqrt(4x+1) is 1/60 (6x-1)(4x+1)^{3/2}+C