What is the integral from 0 to u of (sqrt(t))/(t+1) ?

The integral from 0 to u of (sqrt(t))/(t+1) is 2(-arctan(sqrt(u))+sqrt(u))

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integral from 0 to u of (sqrt(t))/(t+1)

\int_{0}^{u} \frac{t}{t + 1} d t

2 (- arctan (u) + u)

Solution steps

\int_{0}^{u} \frac{t}{t + 1} d t

Apply u-substitution

: \int_{0}^{u} \frac{2 v ^{2}}{v ^{2} + 1} d v

= \int_{0}^{u} \frac{2 v ^{2}}{v ^{2} + 1} d v

Take the constant out:

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

= 2 \cdot \int_{0}^{u} \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= 2 \cdot \int_{0}^{u} - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

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

= 2 (- \int_{0}^{u} \frac{1}{v ^{2} + 1} d v + \int_{0}^{u} 1 d v)

\int_{0}^{u} \frac{1}{v ^{2} + 1} d v = arctan (u)

\int_{0}^{u} 1 d v = u

= 2 (- arctan (u) + u)

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What is the integral from 0 to u of (sqrt(t))/(t+1) ?
The integral from 0 to u of (sqrt(t))/(t+1) is 2(-arctan(sqrt(u))+sqrt(u))