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

\int \frac{1}{z ^{2} + 6 z + 10} d z

\frac{d}{d x} (3 x^{2} + 7 x - 2)

\frac{d x}{d y} = x 1 - y^{2}

\frac{x}{3}

y^{^{''}} - 7 y^{^{'}} + 10 y = e^{3 x}

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