What is the integral of (5x^2)/(x+1) ?

The integral of (5x^2)/(x+1) is 5(((x+1)^2)/2-2(x+1)+ln|x+1|)+C

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integral of (5x^2)/(x+1)

\int \frac{5 x ^{2}}{x + 1} d x

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

Solution steps

\int \frac{5 x ^{2}}{x + 1} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{x + 1} d x

Apply u-substitution

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

= 5 \cdot \int \frac{( u - 1 ) ^{2}}{u} d u

Expand

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

= 5 \cdot \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

= 5 (\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 ∣

= 5 (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

Substitute back

u = x + 1

= 5 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣)

Add a constant to the solution

= 5 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) + C

y = x (1 - x)^{6}

f (s) = s - 3 (s^{2} + 4 s + 13)

y^{2} = 4 x, y^{2} = \frac{4}{x}

n = 1 \sum \infty \frac{( 2 n ) !}{5 ^{n} ( n ! ) ^{2}}

x \to 0 lim (\frac{e ^{x^{5}} - 1}{x})

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