What is the integral of (4+3x)^5 ?

The integral of (4+3x)^5 is 1/18 (4+3x)^6+C

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integral of (4+3x)^5

\int (4 + 3 x)^{5} d x

\frac{1}{18} (4 + 3 x)^{6} + C

Solution steps

\int (4 + 3 x)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{3} d u

= \int \frac{u ^{5}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = 4 + 3 x

= \frac{1}{3} \cdot \frac{( 4 + 3 x ) ^{6}}{6}

Simplify

\frac{1}{3} \cdot \frac{( 4 + 3 x ) ^{6}}{6} : \frac{1}{18} (4 + 3 x)^{6}

= \frac{1}{18} (4 + 3 x)^{6}

Add a constant to the solution

= \frac{1}{18} (4 + 3 x)^{6} + C

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

x^{\frac{2}{3}}, 1

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

(\frac{1}{y ^{2}} - \frac{8}{y ^{4}}) (y + 2 y^{3})

\frac{5}{s ^{2} + 1}