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

The integral of sqrt(4+4t^2+t^4) is t(t^2+2)-(2t^3)/3+C

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

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

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

Solution steps

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

Simplify

t^{2} : tt

= \int 4 + 4 tt + t^{4} d t

Apply Integration By Parts

: t 4 + 4 t^{2} + t^{4} - \int \frac{2 t ^{2} ( t ^{2} + 2 )}{4 + 4 t ^{2} + t ^{4}} d t

= t 4 + 4 t^{2} + t^{4} - \int \frac{2 t ^{2} ( t ^{2} + 2 )}{4 + 4 t ^{2} + t ^{4}} d t

\int \frac{2 t ^{2} ( t ^{2} + 2 )}{4 + 4 t ^{2} + t ^{4}} d t = \frac{2 t ^{3}}{3}

= t 4 + 4 t^{2} + t^{4} - \frac{2 t ^{3}}{3}

4 + 4 t^{2} + t^{4} = t^{2} + 2

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

Add a constant to the solution

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

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What is the integral of sqrt(4+4t^2+t^4) ?
The integral of sqrt(4+4t^2+t^4) is t(t^2+2)-(2t^3)/3+C