What is the integral of (1372x)\sqrt[3]{7x+1} ?

The integral of (1372x)\sqrt[3]{7x+1} is 28(3/7 (7x+1)^{7/3}-3/4 (7x+1)^{4/3})+C

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integral of (1372x)\sqrt[3]{7x+1}

28 (\frac{3}{7} (7 x + 1)^{\frac{7}{3}} - \frac{3}{4} (7 x + 1)^{\frac{4}{3}}) + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

= 1372 \cdot \frac{1}{49} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 7 x + 1

= 1372 \cdot \frac{1}{49} (\frac{3}{7} (7 x + 1)^{\frac{7}{3}} - \frac{3}{4} (7 x + 1)^{\frac{4}{3}})

Simplify

1372 \cdot \frac{1}{49} (\frac{3}{7} (7 x + 1)^{\frac{7}{3}} - \frac{3}{4} (7 x + 1)^{\frac{4}{3}}) : 28 (\frac{3}{7} (7 x + 1)^{\frac{7}{3}} - \frac{3}{4} (7 x + 1)^{\frac{4}{3}})

= 28 (\frac{3}{7} (7 x + 1)^{\frac{7}{3}} - \frac{3}{4} (7 x + 1)^{\frac{4}{3}})

Add a constant to the solution

= 28 (\frac{3}{7} (7 x + 1)^{\frac{7}{3}} - \frac{3}{4} (7 x + 1)^{\frac{4}{3}}) + C

\frac{d}{d x} (ln (arctan (3 x^{3})))

\frac{d}{d x} (\frac{a}{x ^{2} + a ^{2}})

(1 + cos (t)) y^{^{'}} = (1 + e^{- y}) sin (t)

\frac{d}{d x} (lo g_{4} (2 x + 3))

What is the integral of (1372x)\sqrt[3]{7x+1} ?
The integral of (1372x)\sqrt[3]{7x+1} is 28(3/7 (7x+1)^{7/3}-3/4 (7x+1)^{4/3})+C