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

The integral of sqrt(1+x^2(x^2+2)) is (x^3)/3+x+C

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

\int 1 + x^{2} (x^{2} + 2) d x

\frac{x ^{3}}{3} + x + C

Solution steps

\int 1 + x^{2} (x^{2} + 2) d x

Apply Integral Substitution

= \int 2 1 + 2 u^{2} (2 u^{2} + 2) d u

Take the constant out:

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

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

(2 u^{2} + 2) = 2 (u^{2} + 1)

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

Simplify

= 2 \cdot \int 1 + 4 u^{2} (u^{2} + 1) d u

Expand

4 u^{2} (u^{2} + 1) : 4 u^{4} + 4 u^{2}

= 2 \cdot \int 1 + 4 u^{4} + 4 u^{2} d u

Factor

4 u^{4} + 4 u^{2} + 1 : (2 u^{2} + 1)^{2}

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

(2 u^{2} + 1)^{2} = ((2 u^{2} + 1)),

assuming

(2 u^{2} + 1) \geq 0

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

Apply the Sum Rule:

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

= 2 (\int 2 u^{2} d u + \int 1 d u)

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

\int 1 d u = u

= 2 (\frac{2 u ^{3}}{3} + u)

Substitute back

u = \frac{x}{2}

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

Simplify

2 \frac{2 ( \frac{x}{2} ) ^{3}}{3} + \frac{x}{2} : \frac{x ^{3}}{3} + x

= \frac{x ^{3}}{3} + x

Add a constant to the solution

= \frac{x ^{3}}{3} + x + C

What is the integral of sqrt(1+x^2(x^2+2)) ?
The integral of sqrt(1+x^2(x^2+2)) is (x^3)/3+x+C