What is the integral from 0 to 1 of 10sqrt(x^2+1) ?

The integral from 0 to 1 of 10sqrt(x^2+1) is 5(ln(1+sqrt(2))+sqrt(2))

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integral from 0 to 1 of 10sqrt(x^2+1)

\int_{0}^{1} 10 x^{2} + 1 d x

5 (ln (1 + 2) + 2)

Decimal

11.47793 \dots

Solution steps

\int_{0}^{1} 10 x^{2} + 1 d x

Take the constant out:

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

= 10 \cdot \int_{0}^{1} x^{2} + 1 d x

Apply Trigonometric Substitution

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

= 10 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 10 ([\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)

= 10 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

10 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 10 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 10 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 10 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= 5 (ln (1 + 2) + 2)

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What is the integral from 0 to 1 of 10sqrt(x^2+1) ?
The integral from 0 to 1 of 10sqrt(x^2+1) is 5(ln(1+sqrt(2))+sqrt(2))