What is the integral of sqrt(1+x^2)x^5 ?

The integral of sqrt(1+x^2)x^5 is 1/105 (15x^4-12x^2+8)(x^2+1)^{3/2}+C

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integral of sqrt(1+x^2)x^5

\int 1 + x^{2} x^{5} d x

\frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int 1 + x^{2} x^{5} d x

Apply u-substitution

: \int \frac{u ^{2} u + 1}{2} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int (v - 1)^{2} v d v

= \frac{1}{2} \cdot \int (v - 1)^{2} v d v

Expand

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

= \frac{1}{2} \cdot \int v^{\frac{5}{2}} - 2 v^{\frac{3}{2}} + 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

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

\int v^{\frac{5}{2}} d v = \frac{2}{7} v^{\frac{7}{2}}

\int 2 v^{\frac{3}{2}} d v = \frac{4}{5} v^{\frac{5}{2}}

\int v d v = \frac{2}{3} v^{\frac{3}{2}}

= \frac{1}{2} (\frac{2}{7} v^{\frac{7}{2}} - \frac{4}{5} v^{\frac{5}{2}} + \frac{2}{3} v^{\frac{3}{2}})

Substitute back

= \frac{1}{2} (\frac{2}{7} (x^{2} + 1)^{\frac{7}{2}} - \frac{4}{5} (x^{2} + 1)^{\frac{5}{2}} + \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{7} (x^{2} + 1)^{\frac{7}{2}} - \frac{4}{5} (x^{2} + 1)^{\frac{5}{2}} + \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}) : \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{105} (15 x^{4} - 12 x^{2} + 8) (x^{2} + 1)^{\frac{3}{2}} + C

5 + x e^{3 x}

\int \frac{2 sin ( x )}{cos ^{4} ( x )} d x

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

x \to 3 lim (\frac{4 x + 8}{x - 5})

y = e^{x}, at x = 0

What is the integral of sqrt(1+x^2)x^5 ?
The integral of sqrt(1+x^2)x^5 is 1/105 (15x^4-12x^2+8)(x^2+1)^{3/2}+C