What is the integral of 2x(x^2+6)^3 ?

The integral of 2x(x^2+6)^3 is 1/4 (x^2+6)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2x(x^2+6)^3

\int 2 x (x^{2} + 6)^{3} d x

\frac{1}{4} (x^{2} + 6)^{4} + C

Solution steps

\int 2 x (x^{2} + 6)^{3} d x

Take the constant out:

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

= 2 \cdot \int x (x^{2} + 6)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

= 2 \cdot \int \frac{u ^{3}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{2} + 6

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{4}}{4}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{4}}{4} : \frac{1}{4} (x^{2} + 6)^{4}

= \frac{1}{4} (x^{2} + 6)^{4}

Add a constant to the solution

= \frac{1}{4} (x^{2} + 6)^{4} + C

\int \frac{sin ^{2} ( \frac{1}{t} )}{t ^{2}} d t

n = 0 \sum \infty (- 5 x)^{n}

\frac{d}{d t} ((\frac{t ^{3}}{4} - 2) + (\frac{4}{t} - 3) + cos (t - 2))

\int \frac{- 4 x ^{3}}{2 x ^{3} + x ^{2} - 2 x - 1} d x

What is the integral of 2x(x^2+6)^3 ?
The integral of 2x(x^2+6)^3 is 1/4 (x^2+6)^4+C