What is the integral of (sqrt(t)-3)/(sqrt(t)+1) ?

The integral of (sqrt(t)-3)/(sqrt(t)+1) is -8(sqrt(t)+1-ln|sqrt(t)+1|)+t+C

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integral of (sqrt(t)-3)/(sqrt(t)+1)

\int \frac{t - 3}{t + 1} d t

- 8 (t + 1 - ln t + 1) + t + C

Solution steps

\int \frac{t - 3}{t + 1} d t

\frac{t - 3}{t + 1} = - \frac{4}{t + 1} + 1

= \int - \frac{4}{t + 1} + 1 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 \frac{4}{t + 1} d t + \int 1 d t

\int \frac{4}{t + 1} d t = 8 (t + 1 - ln t + 1)

\int 1 d t = t

= - 8 (t + 1 - ln t + 1) + t

Add a constant to the solution

= - 8 (t + 1 - ln t + 1) + t + C

\int x \cdot (ln (x))^{2} d x

\frac{d}{d x} (\frac{x + 1}{x ^{2} - y + 1})

y^{^{''}} - 6 y^{^{'}} + 9 = 0

(- 5) (0.7)

\int \frac{1}{e ^{v}} d v

What is the integral of (sqrt(t)-3)/(sqrt(t)+1) ?
The integral of (sqrt(t)-3)/(sqrt(t)+1) is -8(sqrt(t)+1-ln|sqrt(t)+1|)+t+C