What is the integral from 0 to 1 of 4sqrt(x^2+1) ?

The integral from 0 to 1 of 4sqrt(x^2+1) is 2(ln(1+sqrt(2))+sqrt(2))

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integral from 0 to 1 of 4sqrt(x^2+1)

\int_{0}^{1} 4 x^{2} + 1 d x

2 (ln (1 + 2) + 2)

Decimal

4.59117 \dots

Solution steps

\int_{0}^{1} 4 x^{2} + 1 d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} x^{2} + 1 d x

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= 4 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

= 4 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= 2 (ln (1 + 2) + 2)

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What is the integral from 0 to 1 of 4sqrt(x^2+1) ?
The integral from 0 to 1 of 4sqrt(x^2+1) is 2(ln(1+sqrt(2))+sqrt(2))