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

The integral of (x^2+4)/(x+1) is ((x+1)^2)/2-2(x+1)+5ln|x+1|+C

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integral of (x^2+4)/(x+1)

\int \frac{x ^{2} + 4}{x + 1} d x

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

Solution steps

\int \frac{x ^{2} + 4}{x + 1} d x

Apply u-substitution

: \int \frac{u ^{2} - 2 u + 5}{u} d u

= \int \frac{u ^{2} - 2 u + 5}{u} d u

Expand

\frac{u ^{2} - 2 u + 5}{u} : u - 2 + \frac{5}{u}

= \int u - 2 + \frac{5}{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{5}{u} d u

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

\int 2 d u = 2 u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + 5 ln ∣ u ∣

Substitute back

u = x + 1

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

Add a constant to the solution

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

n = 0 \sum \infty 3 (\frac{x}{4})^{n}

f (x) = \frac{1}{x - 5}, at x = 10

g (x) = x^{2} (1 - 7 x)

\frac{d}{d x} ((\frac{x + 7}{x - 7})^{5})

\frac{d}{d x} ((ln (9))^{x})

What is the integral of (x^2+4)/(x+1) ?
The integral of (x^2+4)/(x+1) is ((x+1)^2)/2-2(x+1)+5ln|x+1|+C