Solve the following integration problems: i) integral (x+1)sqrt(2x+x2)\,dx ii) integral01 t t\,dt

Let $u = 2x+x^2$.
Then, differentiate $u$ with respect to $x$:

So, $du = 2(1+x)\,dx$, which means $(1+x)\,dx = \frac{1}{2}\,du$.

The final answer is $\boxed{\frac{1}{3} (2x+x^2)^{3/2} + C}$.

1. ii) $\int_0^1 t \cosh t\,dt$

Let $u = t$ and $dv = \cosh t\,dt$.
Then, $du = dt$ and $v = \int \cosh t\,dt = \sinh t$.

We know that $\sinh 0 = 0$ and $\cosh 0 = 1$.

Using the definitions $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$:

So, the expression becomes:

The final answer is $\boxed{1 - e^{-1}}$.

2. i) $\int \frac{\sec^2(1/x)}{x^2}\,dx$

Let $u = \frac{1}{x} = x^{-1}$.
Then, differentiate $u$ with respect to $x$:

So, $du = -\frac{1}{x^2}\,dx$, which means $\frac{1}{x^2}\,dx = -du$.

The final answer is $\boxed{-\tan(1/x) + C}$.

2. ii) $\int \frac{\sin(\ln x)}{x}\,dx$

Let $u = \ln x$.
Then, differentiate $u$ with respect to $x$:

So, $du = \frac{1}{x}\,dx$.

The final answer is $\boxed{-\cos(\ln x) + C}$.