What is the integral of (4x+3)^9 ?

The integral of (4x+3)^9 is 1/40 (4x+3)^{10}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x+3)^9

\int (4 x + 3)^{9} d x

\frac{1}{40} (4 x + 3)^{10} + C

Solution steps

\int (4 x + 3)^{9} d x

Apply u-substitution

: \int \frac{u ^{9}}{4} d u

= \int \frac{u ^{9}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= \frac{1}{4} \cdot \frac{u ^{10}}{10}

Substitute back

u = 4 x + 3

= \frac{1}{4} \cdot \frac{( 4 x + 3 ) ^{10}}{10}

Simplify

\frac{1}{4} \cdot \frac{( 4 x + 3 ) ^{10}}{10} : \frac{1}{40} (4 x + 3)^{10}

= \frac{1}{40} (4 x + 3)^{10}

Add a constant to the solution

= \frac{1}{40} (4 x + 3)^{10} + C

y = 1 n (1 n (x^{3}))

\int 7 sin^{4} (x) d x

\frac{d}{d x} (ln (x) cos (t))

\frac{d}{d x} (cot (e^{2 x}))

y^{^{''}} - 6 y^{^{'}} = 0

What is the integral of (4x+3)^9 ?
The integral of (4x+3)^9 is 1/40 (4x+3)^{10}+C