What is the integral from 0 to 1 of 8sqrt(x^2+1) ?

The integral from 0 to 1 of 8sqrt(x^2+1) is 4(ln(1+sqrt(2))+sqrt(2))

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integral from 0 to 1 of 8sqrt(x^2+1)

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

4 (ln (1 + 2) + 2)

Decimal

9.18234 \dots

Solution steps

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

Take the constant out:

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

= 8 \cdot \int_{0}^{1} x^{2} + 1 d x

Apply Trigonometric Substitution

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

= 8 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 8 ([\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)

= 8 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

8 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 8 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 8 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 8 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= 4 (ln (1 + 2) + 2)

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What is the integral from 0 to 1 of 8sqrt(x^2+1) ?
The integral from 0 to 1 of 8sqrt(x^2+1) is 4(ln(1+sqrt(2))+sqrt(2))