What is the integral of 1/(-\frac{1){x^2}+2x}-3 ?

The integral of 1/(-\frac{1){x^2}+2x}-3 is 1/6 ln|-1+2x^3|-3x+C

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integral of 1/(-\frac{1){x^2}+2x}-3

\int \frac{1}{- \frac{1}{x ^{2}} + 2 x} - 3 d x

\frac{1}{6} ln - 1 + 2 x^{3} - 3 x + C

Solution steps

\int \frac{1}{- \frac{1}{x ^{2}} + 2 x} - 3 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

= \int \frac{1}{- \frac{1}{x ^{2}} + 2 x} d x - \int 3 d x

\int \frac{1}{- \frac{1}{x ^{2}} + 2 x} d x = \frac{1}{6} ln - 1 + 2 x^{3}

\int 3 d x = 3 x

= \frac{1}{6} ln - 1 + 2 x^{3} - 3 x

Add a constant to the solution

= \frac{1}{6} ln - 1 + 2 x^{3} - 3 x + C

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

f (x) = x^{3} ln (2 x), at x = 5

x \to 1 lim (1 - (x - 1)^{2})

\frac{\partial}{\partial y} (e^{3 x} tan (y))

\int e^{x} cos (2 x) d x

What is the integral of 1/(-\frac{1){x^2}+2x}-3 ?
The integral of 1/(-\frac{1){x^2}+2x}-3 is 1/6 ln|-1+2x^3|-3x+C