What is the integral from 2 to x^2 of t^2sqrt(t^2+9) ?

The integral from 2 to x^2 of t^2sqrt(t^2+9) is

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integral from 2 to x^2 of t^2sqrt(t^2+9)

\int_{2}^{x^{2}} t^{2} t^{2} + 9 d t

81 - \frac{x ^{2} 9 + x ^{4}}{18} + \frac{13}{9} - \frac{1}{2} ln (- \frac{( 2 - 13 ) x ^{2} + 9 + x ^{4}}{9}) + \frac{2 x ^{6} x ^{4} + 9 + 45 x ^{2} x ^{4} + 9 + 243 ln ( - \frac{( 2 - 13 ) ∣ x ^{2} + x ^{4} + 9 ∣}{9} ) - 106 13}{648}

Solution steps

\int_{2}^{x^{2}} t^{2} t^{2} + 9 d t

Apply Trigonometric Substitution

= \int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} 81 sec^{3} (u) tan^{2} (u) d u

Take the constant out:

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

= 81 \cdot \int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= 81 \cdot \int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= 81 \cdot \int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} - sec^{3} (u) + sec^{5} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 81 (- \int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} sec^{3} (u) d u + \int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} sec^{5} (u) d u)

\int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} sec^{3} (u) d u = \frac{x ^{2} 9 + x ^{4}}{18} - \frac{13}{9} + \frac{1}{2} ln (- \frac{( 2 - 13 ) x ^{2} + 9 + x ^{4}}{9})

\int_{a r c t a n (\frac{2}{3})}^{a r c t a n (\frac{x ^{2}}{3})} sec^{5} (u) d u = \frac{2 x ^{6} x ^{4} + 9 + 45 x ^{2} x ^{4} + 9 + 243 ln ( - \frac{( 2 - 13 ) ∣ x ^{2} + x ^{4} + 9 ∣}{9} ) - 106 13}{648}

= 81 - (\frac{x ^{2} 9 + x ^{4}}{18} - \frac{13}{9} + \frac{1}{2} ln (- \frac{( 2 - 13 ) x ^{2} + 9 + x ^{4}}{9})) + \frac{2 x ^{6} x ^{4} + 9 + 45 x ^{2} x ^{4} + 9 + 243 ln ( - \frac{( 2 - 13 ) ∣ x ^{2} + x ^{4} + 9 ∣}{9} ) - 106 13}{648}

Simplify

= 81 - \frac{x ^{2} 9 + x ^{4}}{18} + \frac{13}{9} - \frac{1}{2} ln (- \frac{( 2 - 13 ) x ^{2} + 9 + x ^{4}}{9}) + \frac{2 x ^{6} x ^{4} + 9 + 45 x ^{2} x ^{4} + 9 + 243 ln ( - \frac{( 2 - 13 ) ∣ x ^{2} + x ^{4} + 9 ∣}{9} ) - 106 13}{648}

What is the integral from 2 to x^2 of t^2sqrt(t^2+9) ?
The integral from 2 to x^2 of t^2sqrt(t^2+9) is