What is the integral of xsqrt(x^4+1) ?

The integral of xsqrt(x^4+1) is 1/2 (1/2 x^2sqrt(1+x^4)+1/2 ln|x^2+sqrt(1+x^4)|)+C

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integral of xsqrt(x^4+1)

\int x x^{4} + 1 d x

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

Solution steps

\int x x^{4} + 1 d x

Simplify

x^{4} : (x^{2})^{2}

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

Apply u-substitution

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

Apply Trigonometric Substitution

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

Apply Integral Reduction

= \frac{1}{2} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

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

Substitute back

= \frac{1}{2} (\frac{sec ^{2} ( arctan ( x ^{2} ) ) sin ( arctan ( x ^{2} ) )}{2} + \frac{1}{2} ln tan (arctan (x^{2})) + sec (arctan (x^{2})))

Simplify

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

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

Add a constant to the solution

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

f (x) = 6 x - x^{2}, at x = 0

f (t) = 80 t + t^{2} - \frac{1}{48} t^{3}

- 4 x + x^{2} + 10, 12 - 5 x, 4 x - 6

\int_{1}^{16} \frac{8}{x} d x

\frac{d}{d x} ((2 + 3 x) (3 - 2 x))

What is the integral of xsqrt(x^4+1) ?
The integral of xsqrt(x^4+1) is 1/2 (1/2 x^2sqrt(1+x^4)+1/2 ln|x^2+sqrt(1+x^4)|)+C