What is the integral of y/((1+y^2)^4) ?

The integral of y/((1+y^2)^4) is -1/(6(1+y^2)^3)+C

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integral of y/((1+y^2)^4)

\int \frac{y}{( 1 + y ^{2} ) ^{4}} d y

- \frac{1}{6 ( 1 + y ^{2} ) ^{3}} + C

Solution steps

\int \frac{y}{( 1 + y ^{2} ) ^{4}} d y

Apply u-substitution

: \int \frac{1}{2 u ^{4}} d u

= \int \frac{1}{2 u ^{4}} 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 \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{2} (- \frac{1}{3 u ^{3}})

Substitute back

u = 1 + y^{2}

= \frac{1}{2} (- \frac{1}{3 ( 1 + y ^{2} ) ^{3}})

Simplify

\frac{1}{2} (- \frac{1}{3 ( 1 + y ^{2} ) ^{3}}) : - \frac{1}{6 ( 1 + y ^{2} ) ^{3}}

= - \frac{1}{6 ( 1 + y ^{2} ) ^{3}}

Add a constant to the solution

= - \frac{1}{6 ( 1 + y ^{2} ) ^{3}} + C

[x + (x + sin^{2} (x))^{3}]^{4}

x \to 2 + lim (\frac{x ^{2} - 4}{x - 2})

\int tan^{5} (4 x) sec^{4} (4 x) d x

x \to - \infty lim (- x^{2})

\frac{1}{( s ^{2} - 2 s + 5 )}

What is the integral of y/((1+y^2)^4) ?
The integral of y/((1+y^2)^4) is -1/(6(1+y^2)^3)+C