# Thread: Antiderivative question

1. ## Antiderivative question

Let F(x) be an antiderivative of (ln x)^3/x. If F(1) = 0, then F(9) =

a. .048
b. .144
c. 5.827
d. 23. 308
e. 1,640.250

2. Originally Posted by frozenflames
Let F(x) be an antiderivative of (ln x)^3/x. If F(1) = 0, then F(9) =

a. .048
b. .144
c. 5.827
d. 23. 308
e. 1,640.250
Find,
$\displaystyle \int \frac{(\ln x)^3}{x}$.
Let $\displaystyle u=\ln x$
then $\displaystyle du=\frac{1}{x}dx$
Express the integral as,
$\displaystyle \int (\ln x)^3\cdot \frac{1}{x}dx$
Thus, we have after substitution,
$\displaystyle \int u^3du=\frac{u^4}{4}+C$
Thus, substituting back,
$\displaystyle \frac{(\ln x)^4}{4}+C$
But $\displaystyle F(1)=0$
Thus,
$\displaystyle \frac{(\ln 1)^4}{4}+C=0$
Evaluate, the left part which gives,
$\displaystyle 0+C=0$
Thus, $\displaystyle C=0$
Thus,$\displaystyle F(x)=\frac{(\ln x)^4}{4}$
Thus, $\displaystyle F(9)=\frac{(\ln 9)^4}{4}\approx 5.83$

$\displaystyle \cal{Q}.\cal{E}.\cal{D}$

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# for x> 0, let f (x)=intlogt/t 1

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