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

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

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integral of (3x^2+4x+2)/(2(y-1))

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

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

Solution steps

\int \frac{3 x ^{2} + 4 x + 2}{2 ( y - 1 )} d x

Take the constant out:

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

= \frac{1}{( - 1 + y ) \cdot 2} \cdot \int 2 + 3 x^{2} + 4 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{( - 1 + y ) \cdot 2} (\int 2 d x + \int 3 x^{2} d x + \int 4 x d x)

\int 2 d x = 2 x

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

\int 4 x d x = 2 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty e^{- \frac{3}{n ^{2}}}

n = 0 \sum \infty (1 + \frac{k}{n})^{n}

\int_{0.5}^{1} \frac{3}{8} (2 - x)^{2} d x

\int_{4}^{5} (t t - 4) d t

\int \frac{1}{ln ( t )} d t

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