What is the integral of (x+7)\sqrt[3]{3-2x} ?

The integral of (x+7)\sqrt[3]{3-2x} is -1/4 (-3/7 (3-2x)^{7/3}+51/4 (3-2x)^{4/3})+C

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integral of (x+7)\sqrt[3]{3-2x}

- \frac{1}{4} (- \frac{3}{7} (3 - 2 x)^{\frac{7}{3}} + \frac{51}{4} (3 - 2 x)^{\frac{4}{3}}) + C

Solution steps

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}}

= - \frac{1}{4} (- \frac{3}{7} u^{\frac{7}{3}} + \frac{51}{4} u^{\frac{4}{3}})

Substitute back

u = 3 - 2 x

= - \frac{1}{4} (- \frac{3}{7} (3 - 2 x)^{\frac{7}{3}} + \frac{51}{4} (3 - 2 x)^{\frac{4}{3}})

Add a constant to the solution

= - \frac{1}{4} (- \frac{3}{7} (3 - 2 x)^{\frac{7}{3}} + \frac{51}{4} (3 - 2 x)^{\frac{4}{3}}) + C

\frac{d}{d x} (ln (\frac{( 2 x + 1 ) ^{2}}{( x - 1 )}))

\frac{( 2 s + 1 )}{s ^{2} + 10 + 25}

tan (3 x), 2 sin (3 x), - \frac{π}{9}, \frac{π}{9}

\int x^{2} 7 + x d x

\int t^{2} 7 t - 5 d t

What is the integral of (x+7)\sqrt[3]{3-2x} ?
The integral of (x+7)\sqrt[3]{3-2x} is -1/4 (-3/7 (3-2x)^{7/3}+51/4 (3-2x)^{4/3})+C