What is the integral of-tsqrt(t-1) ?

The integral of-tsqrt(t-1) is -2/5 (t-1)^{5/2}-2/3 (t-1)^{3/2}+C

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integral of-tsqrt(t-1)

\int - t t - 1 d t

- \frac{2}{5} (t - 1)^{\frac{5}{2}} - \frac{2}{3} (t - 1)^{\frac{3}{2}} + C

Solution steps

\int - t t - 1 d t

Take the constant out:

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

= - \int t t - 1 d t

Apply u-substitution

: \int (u + 1) u d u

= - \int (u + 1) u d u

Expand

(u + 1) u : u^{\frac{3}{2}} + u

= - \int u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= - (\frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = t - 1

= - (\frac{2}{5} (t - 1)^{\frac{5}{2}} + \frac{2}{3} (t - 1)^{\frac{3}{2}})

Simplify

= - \frac{2}{5} (t - 1)^{\frac{5}{2}} - \frac{2}{3} (t - 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{5} (t - 1)^{\frac{5}{2}} - \frac{2}{3} (t - 1)^{\frac{3}{2}} + C

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What is the integral of-tsqrt(t-1) ?
The integral of-tsqrt(t-1) is -2/5 (t-1)^{5/2}-2/3 (t-1)^{3/2}+C