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

The integral of (sqrt(y))/(sqrt(y^3+4)) is 2/3 ln((|y^{3/2}+sqrt(4+y^3)|)/2)+C

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

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

\frac{2}{3} ln \frac{y ^{\frac{3}{2}} + 4 + y ^{3}}{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{6} + 4} d u

= \int \frac{2 u ^{2}}{u ^{6} + 4} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{6} + 4} d u

Apply u-substitution

: \int \frac{1}{3 v ^{2} + 4} d v

= 2 \cdot \int \frac{1}{3 v ^{2} + 4} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 4} d v

Apply Trigonometric Substitution

: \int sec (w) d w

= 2 \cdot \frac{1}{3} \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= 2 \cdot \frac{1}{3} ln ∣ tan (w) + sec (w) ∣

Substitute back

= 2 \cdot \frac{1}{3} ln tan (arctan (\frac{1}{2} (y)^{3})) + sec (arctan (\frac{1}{2} (y)^{3}))

Simplify

2 \cdot \frac{1}{3} ln tan (arctan (\frac{1}{2} (y)^{3})) + sec (arctan (\frac{1}{2} (y)^{3})) : \frac{2}{3} ln \frac{y ^{\frac{3}{2}} + 4 + y ^{3}}{2}

= \frac{2}{3} ln \frac{y ^{\frac{3}{2}} + 4 + y ^{3}}{2}

Add a constant to the solution

= \frac{2}{3} ln \frac{y ^{\frac{3}{2}} + 4 + y ^{3}}{2} + C

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What is the integral of (sqrt(y))/(sqrt(y^3+4)) ?
The integral of (sqrt(y))/(sqrt(y^3+4)) is 2/3 ln((|y^{3/2}+sqrt(4+y^3)|)/2)+C