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

The integral from 1 to x^2 of sqrt(1+t) is (2(1+x^2)^{3/2}-4sqrt(2))/3

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

\int_{1}^{x^{2}} 1 + t d t

\frac{2 ( 1 + x ^{2} ) ^{\frac{3}{2}} - 4 2}{3}

Solution steps

\int_{1}^{x^{2}} 1 + t d t

Apply u-substitution

= \int_{2}^{1 + x^{2}} u d u

Apply the Power Rule

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

Compute the boundaries:

\frac{2}{3} (1 + x^{2})^{\frac{3}{2}} - \frac{4 2}{3}

= \frac{2}{3} (1 + x^{2})^{\frac{3}{2}} - \frac{4 2}{3}

Simplify

= \frac{2 ( 1 + x ^{2} ) ^{\frac{3}{2}} - 4 2}{3}

What is the integral from 1 to x^2 of sqrt(1+t) ?
The integral from 1 to x^2 of sqrt(1+t) is (2(1+x^2)^{3/2}-4sqrt(2))/3