What is the integral of (3x+1)(3x^2+2x)^{3/4} ?

The integral of (3x+1)(3x^2+2x)^{3/4} is 2/7 (3x^2+2x)^{7/4}+C

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integral of (3x+1)(3x^2+2x)^{3/4}

\int (3 x + 1) (3 x^{2} + 2 x)^{\frac{3}{4}} d x

\frac{2}{7} (3 x^{2} + 2 x)^{\frac{7}{4}} + C

Solution steps

\int (3 x + 1) (3 x^{2} + 2 x)^{\frac{3}{4}} d x

Apply u-substitution

: \int \frac{u ^{\frac{3}{4}}}{2} d u

= \int \frac{u ^{\frac{3}{4}}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{\frac{3}{4}} d u

Apply the Power Rule

: \frac{4}{7} u^{\frac{7}{4}}

= \frac{1}{2} \cdot \frac{4}{7} u^{\frac{7}{4}}

Substitute back

u = 3 x^{2} + 2 x

= \frac{1}{2} \cdot \frac{4}{7} (3 x^{2} + 2 x)^{\frac{7}{4}}

Simplify

\frac{1}{2} \cdot \frac{4}{7} (3 x^{2} + 2 x)^{\frac{7}{4}} : \frac{2}{7} (3 x^{2} + 2 x)^{\frac{7}{4}}

= \frac{2}{7} (3 x^{2} + 2 x)^{\frac{7}{4}}

Add a constant to the solution

= \frac{2}{7} (3 x^{2} + 2 x)^{\frac{7}{4}} + C

y = 4 x^{4} + 3 x - 5, (1, 2)

f (x) = \frac{1}{2 + 3 x}, at x = 1

y = - 3 x - 5

\frac{\partial}{\partial x} (- 2 (y + 1) x^{- 3})

sin^{2} (x) - cos^{2} (x)

What is the integral of (3x+1)(3x^2+2x)^{3/4} ?
The integral of (3x+1)(3x^2+2x)^{3/4} is 2/7 (3x^2+2x)^{7/4}+C