What is the integral from 0 to x of t^6sqrt(1+t^7) ?

The integral from 0 to x of t^6sqrt(1+t^7) is (2(x^7+1)^{3/2}-2)/(21)

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integral from 0 to x of t^6sqrt(1+t^7)

\int_{0}^{x} t^{6} 1 + t^{7} d t

\frac{2 ( x ^{7} + 1 ) ^{\frac{3}{2}} - 2}{21}

Solution steps

\int_{0}^{x} t^{6} 1 + t^{7} d t

Apply u-substitution

= \int_{0}^{x^{6}} \frac{u ^{\frac{1}{6}} u ^{\frac{7}{6}} + 1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{x^{6}} u^{\frac{1}{6}} u^{\frac{7}{6}} + 1 d u

Apply u-substitution

= \frac{1}{6} \cdot \int_{0}^{(x^{6})^{\frac{1}{6}}} 6 v^{6} v^{7} + 1 d v

Take the constant out:

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

= \frac{1}{6} \cdot 6 \cdot \int_{0}^{(x^{6})^{\frac{1}{6}}} v^{6} v^{7} + 1 d v

Apply u-substitution

= \frac{1}{6} \cdot 6 \cdot \int_{1}^{(x^{6})^{\frac{7}{6}} + 1} \frac{w}{7} d w

Take the constant out:

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

= \frac{1}{6} \cdot 6 \cdot \frac{1}{7} \cdot \int_{1}^{(x^{6})^{\frac{7}{6}} + 1} w d w

Apply the Power Rule

= \frac{1}{6} \cdot 6 \cdot \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{(x^{6})^{\frac{7}{6}} + 1}

Simplify

\frac{1}{6} \cdot 6 \cdot \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{(x^{6})^{\frac{7}{6}} + 1} : \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{x^{7} + 1}

= \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{x^{7} + 1}

Compute the boundaries:

\frac{2}{3} (x^{7} + 1)^{\frac{3}{2}} - \frac{2}{3}

= \frac{1}{7} (\frac{2}{3} (x^{7} + 1)^{\frac{3}{2}} - \frac{2}{3})

Simplify

= \frac{2 ( x ^{7} + 1 ) ^{\frac{3}{2}} - 2}{21}

What is the integral from 0 to x of t^6sqrt(1+t^7) ?
The integral from 0 to x of t^6sqrt(1+t^7) is (2(x^7+1)^{3/2}-2)/(21)