What is the integral of (3x^2+2)/((x^3+2x+3)^5) ?

The integral of (3x^2+2)/((x^3+2x+3)^5) is -1/(4(x^3+2x+3)^4)+C

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integral of (3x^2+2)/((x^3+2x+3)^5)

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

- \frac{1}{4 ( x ^{3} + 2 x + 3 ) ^{4}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{5}} d u

= \int \frac{1}{u ^{5}} d u

Apply the Power Rule

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

= - \frac{1}{4 u ^{4}}

Substitute back

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

= - \frac{1}{4 ( x ^{3} + 2 x + 3 ) ^{4}}

Add a constant to the solution

= - \frac{1}{4 ( x ^{3} + 2 x + 3 ) ^{4}} + C

y^{^{'}} = \frac{x ^{2} + y ^{2}}{x \cdot y}

\frac{d}{d x} (cos (10 x))

x y^{^{'}} + y^{^{'}} - (x + 1) cos (x) + y = 0

\frac{d}{d x} ((e^{- x} + e^{x})^{3})

\int \frac{x}{( x - 3 )} d x

What is the integral of (3x^2+2)/((x^3+2x+3)^5) ?
The integral of (3x^2+2)/((x^3+2x+3)^5) is -1/(4(x^3+2x+3)^4)+C