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

The integral of sqrt(t^4+2t^2+1) is (t^3)/3+t+C

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

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

\frac{t ^{3}}{3} + t + C

Solution steps

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

Factor

t^{4} + 2 t^{2} + 1 : (t^{2} + 1)^{2}

= \int (t^{2} + 1)^{2} d t

(t^{2} + 1)^{2} = ((t^{2} + 1)),

assuming

(t^{2} + 1) \geq 0

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

Apply the Sum Rule:

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

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

\int t^{2} d t = \frac{t ^{3}}{3}

\int 1 d t = t

= \frac{t ^{3}}{3} + t

Add a constant to the solution

= \frac{t ^{3}}{3} + t + C

\int \frac{1}{( 1 + x ^{2} )} d x

1 2^{9 x}

x \to 2 lim (\frac{x - 2}{∣ x - 2 ∣})

\int sin (2 x) e^{(3 x)} d x

y \to 0 lim (\frac{0}{y ^{2}})

What is the integral of sqrt(t^4+2t^2+1) ?
The integral of sqrt(t^4+2t^2+1) is (t^3)/3+t+C