Integral of x(1+x)^n dx
The solution
The answer (Indefinite)
[src]
// 1 log(1 + x) x*log(1 + x) \
|| ----- + ---------- + ------------ for n = -2|
/ || 1 + x 1 + x 1 + x |
| || |
| n || x - log(1 + x) for n = -1|
| x*(1 + x) dx = C + |< |
| || n 2 n n 2 n |
/ || (1 + x) x *(1 + x) n*x*(1 + x) n*x *(1 + x) |
||- ------------ + ------------ + ------------ + ------------- otherwise |
|| 2 2 2 2 |
\\ 2 + n + 3*n 2 + n + 3*n 2 + n + 3*n 2 + n + 3*n /
$$\int x \left(x + 1\right)^{n}\, dx = C + \begin{cases} \frac{x \log{\left(x + 1 \right)}}{x + 1} + \frac{\log{\left(x + 1 \right)}}{x + 1} + \frac{1}{x + 1} & \text{for}\: n = -2 \\x - \log{\left(x + 1 \right)} & \text{for}\: n = -1 \\\frac{n x^{2} \left(x + 1\right)^{n}}{n^{2} + 3 n + 2} + \frac{n x \left(x + 1\right)^{n}}{n^{2} + 3 n + 2} + \frac{x^{2} \left(x + 1\right)^{n}}{n^{2} + 3 n + 2} - \frac{\left(x + 1\right)^{n}}{n^{2} + 3 n + 2} & \text{otherwise} \end{cases}$$
/ -1/2 + log(2) for n = -2
|
| 1 - log(2) for n = -1
|
< n
| 1 2*n*2
|------------ + ------------ otherwise
| 2 2
\2 + n + 3*n 2 + n + 3*n
$$\begin{cases} - \frac{1}{2} + \log{\left(2 \right)} & \text{for}\: n = -2 \\1 - \log{\left(2 \right)} & \text{for}\: n = -1 \\\frac{2 \cdot 2^{n} n}{n^{2} + 3 n + 2} + \frac{1}{n^{2} + 3 n + 2} & \text{otherwise} \end{cases}$$
=
/ -1/2 + log(2) for n = -2
|
| 1 - log(2) for n = -1
|
< n
| 1 2*n*2
|------------ + ------------ otherwise
| 2 2
\2 + n + 3*n 2 + n + 3*n
$$\begin{cases} - \frac{1}{2} + \log{\left(2 \right)} & \text{for}\: n = -2 \\1 - \log{\left(2 \right)} & \text{for}\: n = -1 \\\frac{2 \cdot 2^{n} n}{n^{2} + 3 n + 2} + \frac{1}{n^{2} + 3 n + 2} & \text{otherwise} \end{cases}$$
Piecewise((-1/2 + log(2), n = -2), (1 - log(2), n = -1), (1/(2 + n^2 + 3*n) + 2*n*2^n/(2 + n^2 + 3*n), True))
Use the examples entering the upper and lower limits of integration.