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

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

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integral of x*1/(x+1)

\int x \cdot \frac{1}{x + 1} d x

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

x \to 5 lim (x^{2} - 6 x + 5)

y^{^{''}} = 3 t cos (9 t)

\frac{\partial}{\partial x} (- 3 x \cdot x y)

f (x) = 1 + 2 x

\frac{\partial}{\partial y} (\frac{x + y}{x y})

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