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

The integral from 0 to 1 of tsqrt(1+t^2+t^4) is (3sqrt(3)-1)/8+3/16 ln(sqrt(3)+2)-3/32 ln(3)

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integral from 0 to 1 of tsqrt(1+t^2+t^4)

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

\frac{3 3 - 1}{8} + \frac{3}{16} ln (3 + 2) - \frac{3}{32} ln (3)

Decimal Notation

0.66845 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{1 + u + u ^{2}}{2} d u

= \int_{0}^{1} \frac{1 + u + u ^{2}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} 1 + u + u^{2} d u

Complete the square

1 + u + u^{2} : (u + \frac{1}{2})^{2} + \frac{3}{4}

= \frac{1}{2} \cdot \int_{0}^{1} (u + \frac{1}{2})^{2} + \frac{3}{4} d u

Apply u-substitution

: \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4 v ^{2} + 3}{2} d v

= \frac{1}{2} \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4 v ^{2} + 3}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} 4 v^{2} + 3 d v

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{3}{2} sec^{3} (w) d w

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{3}{2} sec^{3} (w) d w

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} sec^{3} (w) d w

Apply Integral Reduction

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ([\frac{sec ^{2} ( w ) sin ( w )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} sec (w) d w)

\int_{\frac{π}{6}}^{\frac{π}{3}} sec (w) d w = ln (3 + 2) - \frac{1}{2} ln (3)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ([\frac{sec ^{2} ( w ) sin ( w )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ([\frac{sec ^{2} ( w ) sin ( w )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3))) : \frac{3}{8} ([\frac{1}{2} sec (w) tan (w)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

= \frac{3}{8} ([\frac{1}{2} sec (w) tan (w)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

Compute the boundaries:

3 - \frac{1}{3}

= \frac{3}{8} (3 - \frac{1}{3} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

Simplify

= \frac{3 3 - 1}{8} + \frac{3}{16} ln (3 + 2) - \frac{3}{32} ln (3)

x \to 3 - lim (sin (π x) + 1)

\int e^{2 x + e^{x}} d x

x \to 1 + lim (\frac{x + 10}{x + 2})

2 x y \frac{d y}{d x} = y^{2} - x^{2}

\frac{d y}{d x} = 8 x y + 3 y

What is the integral from 0 to 1 of tsqrt(1+t^2+t^4) ?
The integral from 0 to 1 of tsqrt(1+t^2+t^4) is (3sqrt(3)-1)/8+3/16 ln(sqrt(3)+2)-3/32 ln(3)