What is the integral of (y^3)/(y^2+4) ?

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

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integral of (y^3)/(y^2+4)

\int \frac{y ^{3}}{y ^{2} + 4} d y

\frac{1}{2} (y^{2} + 4 - 4 ln y^{2} + 4) + C

Solution steps

\int \frac{y ^{3}}{y ^{2} + 4} d y

Apply u-substitution

: \int \frac{u - 4}{2 u} d u

= \int \frac{u - 4}{2 u} 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{u - 4}{u} d u

Expand

\frac{u - 4}{u} : 1 - \frac{4}{u}

= \frac{1}{2} \cdot \int 1 - \frac{4}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{2} (u - 4 ln ∣ u ∣)

Substitute back

u = y^{2} + 4

= \frac{1}{2} (y^{2} + 4 - 4 ln y^{2} + 4)

Add a constant to the solution

= \frac{1}{2} (y^{2} + 4 - 4 ln y^{2} + 4) + C

f (x) = 4 tan (x), at x = \frac{π}{4}

\int_{- 2}^{3} \frac{43}{x ^{4}} d x

- 5 x^{3}

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 t) + t e^{4 t}

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

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