Antiderivative question

**frozenflames** · March 18th 2006, 12:05 PM

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

**ThePerfectHacker** · March 18th 2006, 03:43 PM

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,
$\int \frac{(\ln x)^3}{x}$ .
Let $u=\ln x$
then $du=\frac{1}{x}dx$
Express the integral as,
$\int (\ln x)^3\cdot \frac{1}{x}dx$
Thus, we have after substitution,
$\int u^3du=\frac{u^4}{4}+C$
Thus, substituting back,
$\frac{(\ln x)^4}{4}+C$
But $F(1)=0$
Thus,
$\frac{(\ln 1)^4}{4}+C=0$
Evaluate, the left part which gives,
$0+C=0$
Thus, $C=0$
Thus, $F(x)=\frac{(\ln x)^4}{4}$
Thus, $F(9)=\frac{(\ln 9)^4}{4}\approx 5.83$

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

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