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

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

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integral from-1 to 1 of 1/(sqrt(t+1))

\int_{- 1}^{1} \frac{1}{t + 1} d t

22

Decimal Notation

2.82842 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{t + 1} d t

Apply u-substitution

: \int_{0}^{2} 2 d u

= \int_{0}^{2} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{2}

Compute the boundaries:

22

= 22

\frac{d y}{d x} = 42 y x^{6}

\int_{0}^{1} (2 x + y)^{8} d x

f (x) = x^{3} - \frac{3}{2} x^{2} + 4 x + 1

(x - 4) y^{^{'}} + y = x^{2} + 8

3 x^{3}

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