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

The integral of ((x^3+1))/(sqrt(x^4+4x)) is 1/2 sqrt(x^4+4x)+C

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integral of ((x^3+1))/(sqrt(x^4+4x))

\int \frac{( x ^{3} + 1 )}{x ^{4} + 4 x} d x

\frac{1}{2} x^{4} + 4 x + C

Solution steps

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{4} + 4 x

= \frac{1}{4} \cdot 2 (x^{4} + 4 x)^{\frac{1}{2}}

Simplify

\frac{1}{4} \cdot 2 (x^{4} + 4 x)^{\frac{1}{2}} : \frac{1}{2} x^{4} + 4 x

= \frac{1}{2} x^{4} + 4 x

Add a constant to the solution

= \frac{1}{2} x^{4} + 4 x + C

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What is the integral of ((x^3+1))/(sqrt(x^4+4x)) ?
The integral of ((x^3+1))/(sqrt(x^4+4x)) is 1/2 sqrt(x^4+4x)+C