What is the integral of (t^3)/(sqrt(t^4+9)) ?

The integral of (t^3)/(sqrt(t^4+9)) is 1/2 sqrt(t^4+9)+C

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integral of (t^3)/(sqrt(t^4+9))

\int \frac{t ^{3}}{t ^{4} + 9} d t

\frac{1}{2} t^{4} + 9 + C

Solution steps

\int \frac{t ^{3}}{t ^{4} + 9} d t

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = t^{4} + 9

= \frac{1}{4} \cdot 2 (t^{4} + 9)^{\frac{1}{2}}

Simplify

\frac{1}{4} \cdot 2 (t^{4} + 9)^{\frac{1}{2}} : \frac{1}{2} t^{4} + 9

= \frac{1}{2} t^{4} + 9

Add a constant to the solution

= \frac{1}{2} t^{4} + 9 + C

\frac{1}{1 + \frac{1}{5} x ^{3}}

\frac{\partial}{\partial y} (2 x e^{x^{2} + y^{2}})

\frac{\partial}{\partial x} ((x y - 2) e^{- x y})

\frac{( ( 2 x + 5 ) ^{7} )}{14}

x^{4} - 8 x^{2} + 16

What is the integral of (t^3)/(sqrt(t^4+9)) ?
The integral of (t^3)/(sqrt(t^4+9)) is 1/2 sqrt(t^4+9)+C