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

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

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integral from 0 to 1 of 5sqrt(x^2+1)

\int_{0}^{1} 5 x^{2} + 1 d x

\frac{5 ( 2 + ln ( 1 + 2 ) )}{2}

Decimal Notation

5.73896 \dots

Solution steps

\int_{0}^{1} 5 x^{2} + 1 d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} x^{2} + 1 d x

Apply Trigonometric Substitution

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

= 5 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 5 ([\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)

= 5 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

5 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 5 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 5 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 5 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= \frac{5 ( 2 + ln ( 1 + 2 ) )}{2}

\int_{- 1}^{1} cosh^{2} (x) d x

\int_{0}^{3} cos (π x) d x

\int_{0}^{r} (r - \frac{r ^{3}}{r ^{2}}) d r

\int_{0}^{π} sin^{3} (x) d x

\int_{- 1}^{2} 5 x^{2} d x

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