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

The integral of ((2t^2+tsqrt(t)-1))/(t^2) is 2t+2sqrt(t)+1/t+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 t + \frac{1}{t} + 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{1}{t} - \frac{1}{t ^{2}}

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

\int 2 d t = 2 t

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

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

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

Simplify

= 2 t + 2 t + \frac{1}{t}

Add a constant to the solution

= 2 t + 2 t + \frac{1}{t} + C

\int \frac{1}{sin ( y ) + 1} d y

y = x^{3}, x = y^{2} - 2

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x - 5)

\int \frac{7}{( x - 1 ) x ^{2} - 2 x - 48} d x

f (x) = \frac{1}{x ^{4}}

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