What is the integral of (2x)^2(5+3^4)^{1/4} ?

The integral of (2x)^2(5+3^4)^{1/4} is (4*86^{1/4}x^3)/3+C

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integral of (2x)^2(5+3^4)^{1/4}

\int (2 x)^{2} (5 + 3^{4})^{\frac{1}{4}} d x

\frac{4 \cdot 8 6 ^{\frac{1}{4}} x ^{3}}{3} + C

Solution steps

\int (2 x)^{2} (5 + 3^{4})^{\frac{1}{4}} d x

Take the constant out:

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

= (5 + 3^{4})^{\frac{1}{4}} \cdot \int (2 x)^{2} d x

Simplify

= 8 6^{\frac{1}{4}} \cdot \int (2 x)^{2} d x

Simplify

(2 x)^{2} : 4 x^{2}

= 8 6^{\frac{1}{4}} \cdot \int 4 x^{2} d x

Take the constant out:

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

= 8 6^{\frac{1}{4}} \cdot 4 \cdot \int x^{2} d x

Apply the Power Rule

: \frac{x ^{3}}{3}

= 8 6^{\frac{1}{4}} \cdot 4 \cdot \frac{x ^{3}}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 \cdot 8 6 ^{\frac{1}{4}} x ^{3}}{3}

Add a constant to the solution

= \frac{4 \cdot 8 6 ^{\frac{1}{4}} x ^{3}}{3} + C

tan (2 x + 5)

\int 2 + tan^{2} (x) d x

x \to \infty + lim (6 x)

7 x^{2} y^{^{''}} + 15 x y^{^{'}} + y = 48 x + 7

What is the integral of (2x)^2(5+3^4)^{1/4} ?
The integral of (2x)^2(5+3^4)^{1/4} is (4*86^{1/4}x^3)/3+C