What is the integral of te^{1/4 ln(t)} ?

The integral of te^{1/4 ln(t)} is 4/9 t^{9/4}+C

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integral of te^{1/4 ln(t)}

\int t e^{\frac{1}{4} l n (t)} d t

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

Solution steps

\int t e^{\frac{1}{4} l n (t)} d t

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{4} = 4^{- 1}

= \int t e^{4^{- 1} l n (t)} d t

Simplify

t e^{4^{- 1} l n (t)} : t^{\frac{5}{4}}

= \int t^{\frac{5}{4}} d t

Apply the Power Rule

: \frac{4}{9} t^{\frac{9}{4}}

= \frac{4}{9} t^{\frac{9}{4}}

Add a constant to the solution

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

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What is the integral of te^{1/4 ln(t)} ?
The integral of te^{1/4 ln(t)} is 4/9 t^{9/4}+C