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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

Take the constant out:

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

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

Rewrite using trig identities

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

Expand

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

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

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

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

Substitute back

u = arctan (\frac{1}{2} t)

= 2 (- ln tan (arctan (\frac{1}{2} t)) + sec (arctan (\frac{1}{2} t)) + \frac{1}{2} sec (arctan (\frac{1}{2} t)) tan (arctan (\frac{1}{2} t)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} t)) + sec (arctan (\frac{1}{2} t)))

Simplify

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

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

Add a constant to the solution

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

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