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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

\int 2 d u = 2 u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + 2 ln ∣ u ∣

Substitute back

u = x + 1

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

Add a constant to the solution

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

sin (x) cos^{2} (x), 4 - 4 x, 0.69275, 0.92811

\frac{d y}{d x} + \frac{3 y}{x} + 2 = 3 x, y (1) = 1

\frac{d}{d x} (3 x^{2} + 5 x - 2)

\int 2 tan^{5} (4 x) d x

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e - 4 x

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