What is the integral of (y^3)/(1+y^4) ?

The integral of (y^3)/(1+y^4) is 1/4 ln|1+y^4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (y^3)/(1+y^4)

\int \frac{y ^{3}}{1 + y ^{4}} d y

\frac{1}{4} ln 1 + y^{4} + C

Solution steps

\int \frac{y ^{3}}{1 + y ^{4}} d y

Apply u-substitution

: \int \frac{1}{4 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 1 + y^{4}

= \frac{1}{4} ln 1 + y^{4}

Add a constant to the solution

= \frac{1}{4} ln 1 + y^{4} + C

\int \frac{e ^{2 t} + e ^{t}}{e ^{2 t} + 1} d t

\int - \frac{3}{e ^{t}} d t

\int x^{2} (7 x^{3} - 9)^{5} d x

\int \frac{1 - x}{1 + x} d x

\int \frac{t}{t ^{2} + 4} d t

What is the integral of (y^3)/(1+y^4) ?
The integral of (y^3)/(1+y^4) is 1/4 ln|1+y^4|+C