What is the integral of (10)/(sqrt(2x+3)) ?

The integral of (10)/(sqrt(2x+3)) is 10sqrt(2x+3)+C

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integral of (10)/(sqrt(2x+3))

\int \frac{10}{2 x + 3} d x

10 2 x + 3 + C

Solution steps

\int \frac{10}{2 x + 3} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{2 x + 3} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 10 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 10 \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 x + 3

= 10 \cdot \frac{1}{2} \cdot 2 (2 x + 3)^{\frac{1}{2}}

Simplify

10 \cdot \frac{1}{2} \cdot 2 (2 x + 3)^{\frac{1}{2}} : 10 2 x + 3

= 10 2 x + 3

Add a constant to the solution

= 10 2 x + 3 + C

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x y + 1)

\frac{\partial}{\partial x} (ln (3 x^{2} + y^{2} + 4))

\frac{d}{d x} (sec (x) (x - tan (x)))

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x}

x = 6 y^{2}, x = 4 + 2 y^{2}

What is the integral of (10)/(sqrt(2x+3)) ?
The integral of (10)/(sqrt(2x+3)) is 10sqrt(2x+3)+C