What is the integral of sqrt(1+x^4)x^3 ?

The integral of sqrt(1+x^4)x^3 is 1/6 (1+x^4)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(1+x^4)x^3

\int 1 + x^{4} x^{3} d x

\frac{1}{6} (1 + x^{4})^{\frac{3}{2}} + C

Solution steps

\int 1 + x^{4} x^{3} d x

Apply u-substitution

: \int \frac{u}{4} d u

= \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + x^{4}

= \frac{1}{4} \cdot \frac{2}{3} (1 + x^{4})^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (1 + x^{4})^{\frac{3}{2}} : \frac{1}{6} (1 + x^{4})^{\frac{3}{2}}

= \frac{1}{6} (1 + x^{4})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (1 + x^{4})^{\frac{3}{2}} + C

\int 10 sec^{2} (x) tan^{4} (x) d x

\int \frac{4 x + 1}{x ^{2} - 6 x + 13} d x

\int \frac{2}{x ^{2} + 2 x} d x

\int 3 t^{2} ln (t) d t

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

What is the integral of sqrt(1+x^4)x^3 ?
The integral of sqrt(1+x^4)x^3 is 1/6 (1+x^4)^{3/2}+C