What is the integral of ((2t^2+tsqrt(t-1)))/(t^2) ?

The integral of ((2t^2+tsqrt(t-1)))/(t^2) is 2t+2(-arctan(sqrt(t-1))+sqrt(t-1))+C

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integral of ((2t^2+tsqrt(t-1)))/(t^2)

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

2 t + 2 (- arctan (t - 1) + t - 1) + C

Solution steps

\int \frac{2 t ^{2} + t t - 1}{t ^{2}} d t

Expand

\frac{2 t ^{2} + t t - 1}{t ^{2}} : 2 + \frac{t - 1}{t}

= \int 2 + \frac{t - 1}{t} d t

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 d t + \int \frac{t - 1}{t} d t

\int 2 d t = 2 t

\int \frac{t - 1}{t} d t = 2 (- arctan (t - 1) + t - 1)

= 2 t + 2 (- arctan (t - 1) + t - 1)

Add a constant to the solution

= 2 t + 2 (- arctan (t - 1) + t - 1) + C

\frac{d x}{d t} + 8 x = 0

\frac{d}{d x} (4 x (sin (x) + cos (x)))

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

\frac{d y}{d x} = \frac{x - 1}{y ^{3}}

x \to + 2 lim (\frac{( 1 )}{( x + 2 )})

What is the integral of ((2t^2+tsqrt(t-1)))/(t^2) ?
The integral of ((2t^2+tsqrt(t-1)))/(t^2) is 2t+2(-arctan(sqrt(t-1))+sqrt(t-1))+C