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

The integral of ((t-1)/t)^2 is ((t-1)^2)/t-2(ln|t|+1/t)+C

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

integral

(\frac{t - 1}{t})^{2}

\frac{( t - 1 ) ^{2}}{t} - 2 (ln ∣ t ∣ + \frac{1}{t}) + C

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

= \frac{( t - 1 ) ^{2}}{t} - 2 (ln ∣ t ∣ + \frac{1}{t}) + C

\frac{1}{( x + 1 ) ^{3}}

e^{\frac{1}{y}}

sin (- 2 x + 1)

t^{- 4}

\frac{x ^{2}}{( x ^{2} - 3 x + 2 )}

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