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

The integral from 0 to x^2 of sqrt(1+t^2) is (x^2sqrt(x^4+1)+ln|x^2+sqrt(x^4+1)|)/2

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

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

\frac{x ^{2} x ^{4} + 1 + ln x ^{2} + x ^{4} + 1}{2}

Solution steps

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

Apply Trigonometric Substitution

= \int_{0}^{a r c t a n (x^{2})} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x^{2})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (x^{2})} sec (u) d u

\int_{0}^{a r c t a n (x^{2})} sec (u) d u = ln x^{2} + x^{4} + 1

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x^{2})} + \frac{1}{2} ln x^{2} + x^{4} + 1

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x^{2})} + \frac{1}{2} ln x^{2} + x^{4} + 1 : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (x^{2})} + \frac{1}{2} ln x^{2} + x^{4} + 1

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (x^{2})} + \frac{1}{2} ln x^{2} + x^{4} + 1

Compute the boundaries:

\frac{1}{2} x^{2} x^{4} + 1

= \frac{1}{2} x^{2} x^{4} + 1 + \frac{1}{2} ln x^{2} + x^{4} + 1

Simplify

= \frac{x ^{2} x ^{4} + 1 + ln x ^{2} + x ^{4} + 1}{2}

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

\int_{0}^{1} \frac{x ^{2}}{2} d x

\int_{- 3}^{1} \frac{1}{x ^{4}} d x

\int_{0}^{4} x x^{2} + 5 d x

\int_{- 1}^{1} 3 x^{5} d x

What is the integral from 0 to x^2 of sqrt(1+t^2) ?
The integral from 0 to x^2 of sqrt(1+t^2) is (x^2sqrt(x^4+1)+ln|x^2+sqrt(x^4+1)|)/2