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

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

Popular >

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

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

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

Solution steps

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

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

Add a constant to the solution

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

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 8

\int \frac{sin ( c x )}{cos ^{2} ( c x )} d x

\frac{14 x + 8 x ^{2}}{( 7 + 8 x ) ^{2}}

\frac{d}{d t} x + 7 x = 5 cos (2 t)

\int_{1}^{3} x 3^{x^{2}} d x

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