What is the integral from 0 to 1 of 2sqrt(1+t^2) ?

The integral from 0 to 1 of 2sqrt(1+t^2) is sqrt(2)+ln(1+sqrt(2))

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integral from 0 to 1 of 2sqrt(1+t^2)

\int_{0}^{1} 2 1 + t^{2} d t

2 + ln (1 + 2)

Decimal

2.29558 \dots

Solution steps

\int_{0}^{1} 2 1 + t^{2} d t

Take the constant out:

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

= 2 \cdot \int_{0}^{1} 1 + t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

= 2 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 2 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 2 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

2 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 2 [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + ln (1 + 2)

= 2 [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + ln (1 + 2)

Compute the boundaries:

\frac{1}{2}

= 2 \cdot \frac{1}{2} + ln (1 + 2)

Simplify

= 2 + ln (1 + 2)

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What is the integral from 0 to 1 of 2sqrt(1+t^2) ?
The integral from 0 to 1 of 2sqrt(1+t^2) is sqrt(2)+ln(1+sqrt(2))