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

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

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integral of (2x+1)ln(x+1)

\int (2 x + 1) ln (x + 1) d x

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

Solution steps

\int (2 x + 1) ln (x + 1) d x

Expand

(2 x + 1) ln (x + 1) : 2 x ln (x + 1) + ln (x + 1)

= \int 2 x ln (x + 1) + ln (x + 1) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 x ln (x + 1) d x + \int ln (x + 1) d x

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

\int ln (x + 1) d x = x ln (x + 1) + ln (x + 1) - x - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{1}{1 + e})

\int \frac{x}{2 y ^{4}} d x

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\int_{0}^{2} 4 t d t

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