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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xsqrt(1+x^4)

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

Simplify

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec^{3} (v) d v

= \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

\frac{d}{d x} (sin^{2} (x - 3 y))

\int e^{x} 3 - e^{x} d x

\int 3 (x - 1) y^{3} e^{y (1 - x)} d x

\int + e^{- 0.1 x} d x

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

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