What is the integral of (2100)/(sqrt(3x+1)) ?

The integral of (2100)/(sqrt(3x+1)) is 1400sqrt(3x+1)+C

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integral of (2100)/(sqrt(3x+1))

\int \frac{2100}{3 x + 1} d x

1400 3 x + 1 + C

Solution steps

\int \frac{2100}{3 x + 1} d x

Take the constant out:

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

= 2100 \cdot \int \frac{1}{3 x + 1} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= 2100 \cdot \int \frac{1}{3 u} d u

Take the constant out:

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

= 2100 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 2100 \cdot \frac{1}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 3 x + 1

= 2100 \cdot \frac{1}{3} \cdot 2 (3 x + 1)^{\frac{1}{2}}

Simplify

2100 \cdot \frac{1}{3} \cdot 2 (3 x + 1)^{\frac{1}{2}} : 1400 3 x + 1

= 1400 3 x + 1

Add a constant to the solution

= 1400 3 x + 1 + C

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What is the integral of (2100)/(sqrt(3x+1)) ?
The integral of (2100)/(sqrt(3x+1)) is 1400sqrt(3x+1)+C