What is the integral of sqrt(4x^2+9x^4) ?

The integral of sqrt(4x^2+9x^4) is 1/27 (4+9x^2)^{3/2}+C

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integral of sqrt(4x^2+9x^4)

\int 4 x^{2} + 9 x^{4} d x

\frac{1}{27} (4 + 9 x^{2})^{\frac{3}{2}} + C

Solution steps

\int 4 x^{2} + 9 x^{4} d x

Factor

4 x^{2} + 9 x^{4} : x^{2} 4 + 9 x^{2}

= \int x^{2} 4 + 9 x^{2} d x

x^{2} = (x),

assuming

x \geq 0

= \int x 4 + 9 x^{2} d x

Apply u-substitution

: \int \frac{u}{18} d u

= \int \frac{u}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{18} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 + 9 x^{2}

= \frac{1}{18} \cdot \frac{2}{3} (4 + 9 x^{2})^{\frac{3}{2}}

Simplify

\frac{1}{18} \cdot \frac{2}{3} (4 + 9 x^{2})^{\frac{3}{2}} : \frac{1}{27} (4 + 9 x^{2})^{\frac{3}{2}}

= \frac{1}{27} (4 + 9 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{27} (4 + 9 x^{2})^{\frac{3}{2}} + C

\frac{d}{d x} (1 0^{n x})

f (x) = \frac{14}{( x + 4 ) ^{3}}

\frac{d y}{d x} = \frac{cos ( x )}{y}

n = 1 \sum \infty \frac{5 n}{2 n - 1}

\int t k d t

What is the integral of sqrt(4x^2+9x^4) ?
The integral of sqrt(4x^2+9x^4) is 1/27 (4+9x^2)^{3/2}+C