What is the integral of-(4ln(t))/(t^2) ?

The integral of-(4ln(t))/(t^2) is -4(-(ln(t))/t-1/t)+C

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integral of-(4ln(t))/(t^2)

\int - \frac{4 ln ( t )}{t ^{2}} d t

- 4 (- \frac{ln ( t )}{t} - \frac{1}{t}) + C

Solution steps

\int - \frac{4 ln ( t )}{t ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 4 \cdot \int \frac{ln ( t )}{t ^{2}} d t

Apply Integration By Parts

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

= - 4 (- \frac{ln ( t )}{t} - \int - \frac{1}{t ^{2}} d t)

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

= - 4 (- \frac{ln ( t )}{t} - \frac{1}{t})

Add a constant to the solution

= - 4 (- \frac{ln ( t )}{t} - \frac{1}{t}) + C

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What is the integral of-(4ln(t))/(t^2) ?
The integral of-(4ln(t))/(t^2) is -4(-(ln(t))/t-1/t)+C