What is the integral of 8x(4x^2+7)^6 ?

The integral of 8x(4x^2+7)^6 is 1/7 (4x^2+7)^7+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8x(4x^2+7)^6

\int 8 x (4 x^{2} + 7)^{6} d x

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

Solution steps

\int 8 x (4 x^{2} + 7)^{6} d x

Take the constant out:

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

= 8 \cdot \int x (4 x^{2} + 7)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{8} d u

= 8 \cdot \int \frac{u ^{6}}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 8 \cdot \frac{1}{8} \cdot \frac{u ^{7}}{7}

Substitute back

u = 4 x^{2} + 7

= 8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 7 ) ^{7}}{7}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 7 ) ^{7}}{7} : \frac{1}{7} (4 x^{2} + 7)^{7}

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

Add a constant to the solution

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

\int \frac{e ^{x}}{2 ^{x}} d x

\int x \cdot a \cdot e^{- a x} d x

\int cos (x) e^{5 + 2 s i n (x)} d x

\int e^{8 x} 8 d x

\int (1 + 6 x)^{4} d x

What is the integral of 8x(4x^2+7)^6 ?
The integral of 8x(4x^2+7)^6 is 1/7 (4x^2+7)^7+C