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

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

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

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

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

Solution steps

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

Simplify

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

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

t^{2} = (t),

assuming

t \geq 0

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = t^{2} + 1

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

Simplify

\frac{( t ^{2} + 1 ) ^{3}}{3} : \frac{1}{3} (t^{2} + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

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