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

The integral of (x^2+1)/((x+1)2) is 1/4 (x^2-2x+4ln|x+1|-3)+C

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

integral

\frac{x ^{2} + 1}{( x + 1 ) 2}

\frac{1}{4} (x^{2} - 2 x + 4 ln ∣ x + 1 ∣ - 3) + C

Solution steps

\int \frac{x ^{2} + 1}{( x + 1 ) \cdot 2} d x

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

= \frac{1}{2} \cdot \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

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

Substitute back

u = x + 1

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

Simplify

\frac{1}{2} (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + 2 ln ∣ x + 1 ∣) : \frac{1}{4} (x^{2} - 2 x + 4 ln ∣ x + 1 ∣ - 3)

= \frac{1}{4} (x^{2} - 2 x + 4 ln ∣ x + 1 ∣ - 3)

Add a constant to the solution

= \frac{1}{4} (x^{2} - 2 x + 4 ln ∣ x + 1 ∣ - 3) + C

\frac{81 x ^{2}}{9 x - 2}

4 cos^{2} (x) - 2 sin^{2} (x)

tan^{2} (3 x - 2)

2 sin^{2} (x)

\frac{3}{( x + 2 )}

What is the integral of (x^2+1)/((x+1)2) ?
The integral of (x^2+1)/((x+1)2) is 1/4 (x^2-2x+4ln|x+1|-3)+C