What is the integral of (2x)/(sqrt(x^2+4)) ?

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

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integral of (2x)/(sqrt(x^2+4))

\int \frac{2 x}{x ^{2} + 4} d x

2 x^{2} + 4 + C

Solution steps

\int \frac{2 x}{x ^{2} + 4} d x

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 4

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

Simplify

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

= 2 x^{2} + 4

Add a constant to the solution

= 2 x^{2} + 4 + C

T \to \infty lim (f (T))

\int sec^{5} (3 x) tan (3 x) d x

\frac{x}{y}

\int \frac{1}{( x ^{2} - 144 ) ^{\frac{3}{2}}} d x

\frac{d y}{d x} = 6 y^{\frac{2}{3}}

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