What is the integral of (3r^2)/(r+7) ?

The integral of (3r^2)/(r+7) is 3(((r+7)^2)/2-14(r+7)+49ln|r+7|)+C

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integral of (3r^2)/(r+7)

\int \frac{3 r ^{2}}{r + 7} d r

3 (\frac{( r + 7 ) ^{2}}{2} - 14 (r + 7) + 49 ln ∣ r + 7 ∣) + C

Solution steps

\int \frac{3 r ^{2}}{r + 7} d r

Take the constant out:

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

= 3 \cdot \int \frac{r ^{2}}{r + 7} d r

Apply u-substitution

: \int \frac{( u - 7 ) ^{2}}{u} d u

= 3 \cdot \int \frac{( u - 7 ) ^{2}}{u} d u

Expand

\frac{( u - 7 ) ^{2}}{u} : u - 14 + \frac{49}{u}

= 3 \cdot \int u - 14 + \frac{49}{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

= 3 (\int u d u - \int 14 d u + \int \frac{49}{u} d u)

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

\int 14 d u = 14 u

\int \frac{49}{u} d u = 49 ln ∣ u ∣

= 3 (\frac{u ^{2}}{2} - 14 u + 49 ln ∣ u ∣)

Substitute back

u = r + 7

= 3 (\frac{( r + 7 ) ^{2}}{2} - 14 (r + 7) + 49 ln ∣ r + 7 ∣)

Add a constant to the solution

= 3 (\frac{( r + 7 ) ^{2}}{2} - 14 (r + 7) + 49 ln ∣ r + 7 ∣) + C

\frac{s}{s ^{2} - 2 s - 3}

y = 4 + 9 x

y^{^{'}} = y - y^{3}

sin (3 t) cos (5 t)

(2 x - 1) d x + (5 y + 8) d y = 0

What is the integral of (3r^2)/(r+7) ?
The integral of (3r^2)/(r+7) is 3(((r+7)^2)/2-14(r+7)+49ln|r+7|)+C