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

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

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

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

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

Solution steps

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

Simplify

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

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

t^{2} = (t),

assuming

t \geq 0

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

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

Simplify

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

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

Add a constant to the solution

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

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