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

The integral of t^2sqrt(1+t^2) is -1/4 (1/2 tsqrt(1+t^2)+1/2 ln|t+sqrt(1+t^2)|)+1/4 t(t^2+1)^{3/2}+C

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integral of t^2sqrt(1+t^2)

\int t^{2} 1 + t^{2} d t

- \frac{1}{4} (\frac{1}{2} t 1 + t^{2} + \frac{1}{2} ln t + 1 + t^{2}) + \frac{1}{4} t (t^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int t^{2} 1 + t^{2} d t

Apply Trigonometric Substitution

: \int sec^{3} (u) tan^{2} (u) d u

= \int sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

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

= \int - 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

= - \int sec^{3} (u) d u + \int sec^{5} (u) d u

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= - (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (t)

= - (\frac{1}{2} sec (arctan (t)) tan (arctan (t)) + \frac{1}{2} ln ∣ tan (arctan (t)) + sec (arctan (t)) ∣) + \frac{sec ^{4} ( arctan ( t ) ) sin ( arctan ( t ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (t)) tan (arctan (t)) + \frac{1}{2} ln ∣ tan (arctan (t)) + sec (arctan (t)) ∣)

Simplify

- (\frac{1}{2} sec (arctan (t)) tan (arctan (t)) + \frac{1}{2} ln ∣ tan (arctan (t)) + sec (arctan (t)) ∣) + \frac{sec ^{4} ( arctan ( t ) ) sin ( arctan ( t ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (t)) tan (arctan (t)) + \frac{1}{2} ln ∣ tan (arctan (t)) + sec (arctan (t)) ∣) : - \frac{1}{4} (\frac{1}{2} t 1 + t^{2} + \frac{1}{2} ln t + 1 + t^{2}) + \frac{1}{4} t (t^{2} + 1)^{\frac{3}{2}}

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

Add a constant to the solution

= - \frac{1}{4} (\frac{1}{2} t 1 + t^{2} + \frac{1}{2} ln t + 1 + t^{2}) + \frac{1}{4} t (t^{2} + 1)^{\frac{3}{2}} + C

\frac{d}{d x} (- x^{2} + 9)

\frac{7 e ^{x}}{( 7 + e ^{x} ) ^{2}}

\frac{d}{d t} (t^{r})

\int \frac{1}{x ^{2} + 10} d x

\int \frac{2}{- x ^{2} + 5} d x

What is the integral of t^2sqrt(1+t^2) ?
The integral of t^2sqrt(1+t^2) is -1/4 (1/2 tsqrt(1+t^2)+1/2 ln|t+sqrt(1+t^2)|)+1/4 t(t^2+1)^{3/2}+C