What is the integral of (x^2+2x)/(sqrt(x^3+3x^2+2)) ?

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

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integral of (x^2+2x)/(sqrt(x^3+3x^2+2))

\int \frac{x ^{2} + 2 x}{x ^{3} + 3 x ^{2} + 2} d x

\frac{2}{3} x^{3} + 3 x^{2} + 2 + C

Solution steps

\int \frac{x ^{2} + 2 x}{x ^{3} + 3 x ^{2} + 2} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{3} + 3 x^{2} + 2

= \frac{1}{3} \cdot 2 (x^{3} + 3 x^{2} + 2)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot 2 (x^{3} + 3 x^{2} + 2)^{\frac{1}{2}} : \frac{2}{3} x^{3} + 3 x^{2} + 2

= \frac{2}{3} x^{3} + 3 x^{2} + 2

Add a constant to the solution

= \frac{2}{3} x^{3} + 3 x^{2} + 2 + C

y^{^{''}} - y = t^{2} e^{t}

\int \frac{x ^{2}}{x ^{3}} d x

\frac{d y}{( 6 + y ) ^{2}} = \frac{d x}{( 10 + 9 x ) ^{4}}

\frac{df}{d θ} = sin (θ) + csc (θ)

\int (2 x^{3} - 5 x + 7) d x

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