What is the integral of sqrt(7x+6) ?

The integral of sqrt(7x+6) is 2/21 (7x+6)^{3/2}+C

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integral of sqrt(7x+6)

\int 7 x + 6 d x

\frac{2}{21} (7 x + 6)^{\frac{3}{2}} + C

Solution steps

\int 7 x + 6 d x

Apply u-substitution

: \int \frac{u}{7} d u

= \int \frac{u}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{7} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 7 x + 6

= \frac{1}{7} \cdot \frac{2}{3} (7 x + 6)^{\frac{3}{2}}

Simplify

\frac{1}{7} \cdot \frac{2}{3} (7 x + 6)^{\frac{3}{2}} : \frac{2}{21} (7 x + 6)^{\frac{3}{2}}

= \frac{2}{21} (7 x + 6)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{21} (7 x + 6)^{\frac{3}{2}} + C

x y

\frac{s - 30}{s ( s ^{2} + 4 s + 29 )}

\int x^{4} d x

y = 5 x^{3} - 4 x^{2} + 3 x - 6

\frac{\partial}{\partial x} (2 x y arcsin (yz))

What is the integral of sqrt(7x+6) ?
The integral of sqrt(7x+6) is 2/21 (7x+6)^{3/2}+C