Integral of sqrt(1+x) dx

1. The integral of a constant times a function is the constant times the integral of the function:

   $$\int t \left(x + 1\right)^{2}\, dx = t \int \left(x + 1\right)^{2}\, dx$$

   1. There are multiple ways to do this integral.

      Method #1

      1. Let $u = x + 1$.

         Then let $du = dx$ and substitute $du$:

         $$\int u^{2}\, du$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

         Now substitute $u$ back in:

         $$\frac{\left(x + 1\right)^{3}}{3}$$

      Method #2

      1. Rewrite the integrand:

         $$\left(x + 1\right)^{2} = x^{2} + 2 x + 1$$

      2. Integrate term-by-term:

         1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int 2 x\, dx = 2 \int x\, dx$$

            1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int x\, dx = \frac{x^{2}}{2}$$

            So, the result is: $x^{2}$

         1. The integral of a constant is the constant times the variable of integration:

            $$\int 1\, dx = x$$

         The result is: $\frac{x^{3}}{3} + x^{2} + x$

   So, the result is: $\frac{t \left(x + 1\right)^{3}}{3}$

2. Add the constant of integration:

   $$\frac{t \left(x + 1\right)^{3}}{3}+ \mathrm{constant}$$

The answer is: