What is the integral of t^2sqrt(1+4t^2) ?

The integral of t^2sqrt(1+4t^2) is 1/64 (-2tsqrt(1+4t^2)-ln|2t+sqrt(1+4t^2)|+4t(4t^2+1)^{3/2})+C

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integral of t^2sqrt(1+4t^2)

\int t^{2} 1 + 4 t^{2} d t

\frac{1}{64} (- 2 t 1 + 4 t^{2} - ln 2 t + 1 + 4 t^{2} + 4 t (4 t^{2} + 1)^{\frac{3}{2}}) + C

Solution steps

\int t^{2} 1 + 4 t^{2} d t

Apply Trigonometric Substitution

: \int \frac{1}{8} sec^{3} (u) tan^{2} (u) d u

= \int \frac{1}{8} 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

= \frac{1}{8} \cdot \int sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

Expand

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

= \frac{1}{8} \cdot \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

= \frac{1}{8} (- \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}{8} (- (\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 (2 t)

= \frac{1}{8} (- (\frac{1}{2} sec (arctan (2 t)) tan (arctan (2 t)) + \frac{1}{2} ln ∣ tan (arctan (2 t)) + sec (arctan (2 t)) ∣) + \frac{sec ^{4} ( arctan ( 2 t ) ) sin ( arctan ( 2 t ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (2 t)) tan (arctan (2 t)) + \frac{1}{2} ln ∣ tan (arctan (2 t)) + sec (arctan (2 t)) ∣))

Simplify

\frac{1}{8} (- (\frac{1}{2} sec (arctan (2 t)) tan (arctan (2 t)) + \frac{1}{2} ln ∣ tan (arctan (2 t)) + sec (arctan (2 t)) ∣) + \frac{sec ^{4} ( arctan ( 2 t ) ) sin ( arctan ( 2 t ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (2 t)) tan (arctan (2 t)) + \frac{1}{2} ln ∣ tan (arctan (2 t)) + sec (arctan (2 t)) ∣)) : \frac{1}{64} (- 2 t 1 + 4 t^{2} - ln 2 t + 1 + 4 t^{2} + 4 t (4 t^{2} + 1)^{\frac{3}{2}})

= \frac{1}{64} (- 2 t 1 + 4 t^{2} - ln 2 t + 1 + 4 t^{2} + 4 t (4 t^{2} + 1)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{64} (- 2 t 1 + 4 t^{2} - ln 2 t + 1 + 4 t^{2} + 4 t (4 t^{2} + 1)^{\frac{3}{2}}) + C

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What is the integral of t^2sqrt(1+4t^2) ?
The integral of t^2sqrt(1+4t^2) is 1/64 (-2tsqrt(1+4t^2)-ln|2t+sqrt(1+4t^2)|+4t(4t^2+1)^{3/2})+C