What is the integral of (8x^7+x^5)sqrt(6x^8+x^6) ?

The integral of (8x^7+x^5)sqrt(6x^8+x^6) is 1/9 x^9(6x^2+1)^{3/2}+C

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integral of (8x^7+x^5)sqrt(6x^8+x^6)

\int (8 x^{7} + x^{5}) 6 x^{8} + x^{6} d x

\frac{1}{9} x^{9} (6 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int (8 x^{7} + x^{5}) 6 x^{8} + x^{6} d x

Apply u-substitution

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

= \int \frac{u}{6} d u

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 6 x^{8} + x^{6}

= \frac{1}{6} \cdot \frac{2}{3} (6 x^{8} + x^{6})^{\frac{3}{2}}

Simplify

\frac{1}{6} \cdot \frac{2}{3} (6 x^{8} + x^{6})^{\frac{3}{2}} : \frac{1}{9} x^{9} (6 x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{9} x^{9} (6 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9} x^{9} (6 x^{2} + 1)^{\frac{3}{2}} + C

\int \frac{sinh ( x )}{1 - 4 cosh ( x )} d x

x \to 0 lim ((x - 1)^{\frac{1}{2}})

(2 x y^{2} + 3 x^{2} y) d x + (2 x^{2} y + x^{3}) d y = 0

(- 5, 3), (5, - 3)

f (x) = x^{2} + 4

What is the integral of (8x^7+x^5)sqrt(6x^8+x^6) ?
The integral of (8x^7+x^5)sqrt(6x^8+x^6) is 1/9 x^9(6x^2+1)^{3/2}+C