What is the integral of (m^2)/(m+1) ?

The integral of (m^2)/(m+1) is ((m+1)^2)/2-2(m+1)+ln|m+1|+C

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integral of (m^2)/(m+1)

\int \frac{m ^{2}}{m + 1} d m

\frac{( m + 1 ) ^{2}}{2} - 2 (m + 1) + ln ∣ m + 1 ∣ + C

Solution steps

\int \frac{m ^{2}}{m + 1} d m

Apply u-substitution

= \int \frac{( u - 1 ) ^{2}}{u} d u

Expand

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= \int u - 2 + \frac{1}{u} d u

Apply the Sum Rule:

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

= \int u d u - \int 2 d u + \int \frac{1}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + ln ∣ u ∣

Substitute back

u = m + 1

= \frac{( m + 1 ) ^{2}}{2} - 2 (m + 1) + ln ∣ m + 1 ∣

Add a constant to the solution

= \frac{( m + 1 ) ^{2}}{2} - 2 (m + 1) + ln ∣ m + 1 ∣ + C

\frac{d}{d x} (\frac{x + 10}{x - 10})

\int x 1 + x d x

d er i v a t i v e ln (8 x + 5)

x y^{^{'}} + 9 y = 12 y^{2}, y (1) = - 1

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

What is the integral of (m^2)/(m+1) ?
The integral of (m^2)/(m+1) is ((m+1)^2)/2-2(m+1)+ln|m+1|+C