What is the integral from 0 to 1 of 3sqrt(x^2+1) ?

The integral from 0 to 1 of 3sqrt(x^2+1) is (3(sqrt(2)+ln(1+sqrt(2))))/2

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integral from 0 to 1 of 3sqrt(x^2+1)

\int_{0}^{1} 3 x^{2} + 1 d x

\frac{3 ( 2 + ln ( 1 + 2 ) )}{2}

Decimal

3.44338 \dots

Solution steps

\int_{0}^{1} 3 x^{2} + 1 d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} x^{2} + 1 d x

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= 3 ([\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)

= 3 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

3 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 3 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 3 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 3 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= \frac{3 ( 2 + ln ( 1 + 2 ) )}{2}

2 + 7 t

y^{^{'}} + 2 x^{5} y = \frac{x ^{5}}{y ^{2}}

\frac{10 x}{- 3 y ^{2}}

x \to 0 + lim (\frac{ln ( x )}{x ^{3}})

\frac{x}{x + 2}

What is the integral from 0 to 1 of 3sqrt(x^2+1) ?
The integral from 0 to 1 of 3sqrt(x^2+1) is (3(sqrt(2)+ln(1+sqrt(2))))/2