What is the integral of 6x^5sqrt(3x^4+2) ?

The integral of 6x^5sqrt(3x^4+2) is

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integral of 6x^5sqrt(3x^4+2)

\int 6 x^{5} 3 x^{4} + 2 d x

- \frac{1}{4} x^{2} 2 + 3 x^{4} - \frac{1}{2 3} ln \frac{3}{2} x^{2} + \frac{1}{2} (2 + 3 x^{4}) + \frac{1}{4} x^{2} (3 x^{4} + 2)^{\frac{3}{2}} + C

Solution steps

\int 6 x^{5} 3 x^{4} + 2 d x

Take the constant out:

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

= 6 \cdot \int x^{5} 3 x^{4} + 2 d x

Apply u-substitution

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

= 6 \cdot \int \frac{u ^{2} 3 u ^{2} + 2}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int u^{2} 3 u^{2} + 2 d u

Apply Trigonometric Substitution

: \int \frac{4}{3 3} sec^{3} (v) tan^{2} (v) d v

= 6 \cdot \frac{1}{2} \cdot \int \frac{4}{3 3} sec^{3} (v) tan^{2} (v) d v

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \frac{4}{3 3} \cdot \int sec^{3} (v) tan^{2} (v) d v

Rewrite using trig identities

= 6 \cdot \frac{1}{2} \cdot \frac{4}{3 3} \cdot \int sec^{3} (v) (- 1 + sec^{2} (v)) d v

Expand

sec^{3} (v) (- 1 + sec^{2} (v)) : - sec^{3} (v) + sec^{5} (v)

= 6 \cdot \frac{1}{2} \cdot \frac{4}{3 3} \cdot \int - sec^{3} (v) + sec^{5} (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 \cdot \frac{1}{2} \cdot \frac{4}{3 3} (- \int sec^{3} (v) d v + \int sec^{5} (v) d v)

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

\int sec^{5} (v) d v = \frac{sec ^{4} ( v ) sin ( v )}{4} + \frac{3}{4} (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

= 6 \cdot \frac{1}{2} \cdot \frac{4}{3 3} (- (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣) + \frac{sec ^{4} ( v ) sin ( v )}{4} + \frac{3}{4} (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣))

Substitute back

= 6 \cdot \frac{1}{2} \cdot \frac{4}{3 3} - (\frac{1}{2} sec (arctan (\frac{3}{2} x^{2})) tan (arctan (\frac{3}{2} x^{2})) + \frac{1}{2} ln tan (arctan (\frac{3}{2} x^{2})) + sec (arctan (\frac{3}{2} x^{2}))) + \frac{sec ^{4} ( arctan ( \frac{3}{2} x ^{2} ) ) sin ( arctan ( \frac{3}{2} x ^{2} ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{3}{2} x^{2})) tan (arctan (\frac{3}{2} x^{2})) + \frac{1}{2} ln tan (arctan (\frac{3}{2} x^{2})) + sec (arctan (\frac{3}{2} x^{2})))

Simplify

6 \cdot \frac{1}{2} \cdot \frac{4}{3 3} - (\frac{1}{2} sec (arctan (\frac{3}{2} x^{2})) tan (arctan (\frac{3}{2} x^{2})) + \frac{1}{2} ln tan (arctan (\frac{3}{2} x^{2})) + sec (arctan (\frac{3}{2} x^{2}))) + \frac{sec ^{4} ( arctan ( \frac{3}{2} x ^{2} ) ) sin ( arctan ( \frac{3}{2} x ^{2} ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{3}{2} x^{2})) tan (arctan (\frac{3}{2} x^{2})) + \frac{1}{2} ln tan (arctan (\frac{3}{2} x^{2})) + sec (arctan (\frac{3}{2} x^{2}))) : - \frac{1}{4} x^{2} 2 + 3 x^{4} - \frac{1}{2 3} ln \frac{3}{2} x^{2} + \frac{1}{2} (2 + 3 x^{4}) + \frac{1}{4} x^{2} (3 x^{4} + 2)^{\frac{3}{2}}

= - \frac{1}{4} x^{2} 2 + 3 x^{4} - \frac{1}{2 3} ln \frac{3}{2} x^{2} + \frac{1}{2} (2 + 3 x^{4}) + \frac{1}{4} x^{2} (3 x^{4} + 2)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{4} x^{2} 2 + 3 x^{4} - \frac{1}{2 3} ln \frac{3}{2} x^{2} + \frac{1}{2} (2 + 3 x^{4}) + \frac{1}{4} x^{2} (3 x^{4} + 2)^{\frac{3}{2}} + C

y = 9 x, y = 70 - x^{2}

\int (\frac{5}{x}) d x

\frac{\partial}{\partial y} (3 x^{2} y - y^{3} + x - y)

\frac{d}{d x} (\frac{1}{27} (x^{5} + 2 x^{3}))

\int 26 x - x^{2} d x