What is the integral of (4x^2+1)^3*8x ?

The integral of (4x^2+1)^3*8x is 1/4 (4x^2+1)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^2+1)^3*8x

\int (4 x^{2} + 1)^{3} \cdot 8 x d x

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

Solution steps

\int (4 x^{2} + 1)^{3} \cdot 8 x d x

Take the constant out:

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

= 8 \cdot \int (4 x^{2} + 1)^{3} x d x

Apply u-substitution

: \int \frac{u ^{3}}{8} d u

= 8 \cdot \int \frac{u ^{3}}{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^{3} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{8} \cdot \frac{u ^{4}}{4}

Substitute back

u = 4 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

x (1 - x^{2})

x \to 2 lim (x x^{2} - 4)

y^{^{''}} + 5 y^{^{'}} + y = 0

\int \frac{( 6 x ^{2} + 4 )}{( 3 y ^{2} - 4 )}

\frac{\partial}{\partial x} (4 x^{8} y^{6} + 5 x^{7} y^{5})

What is the integral of (4x^2+1)^3*8x ?
The integral of (4x^2+1)^3*8x is 1/4 (4x^2+1)^4+C