Evaluating a PDF to find an unkown constant

Question

Evaluate the PDF to find K

$1 = \int_0^{\infty} \frac{k}{(1+x)^4}$

using u substitution I got:

$u=1+x$ and $du= dx$

$\int_0^{\infty} \frac{k}{u^4} \to \frac{k}{\frac{-1}{3} u^3}$

Since $u^4 du = \frac{u^{-4+1}}{-4+1}=$ $\frac{u^{-3}}{-3}$

which evaluating gives:

$1=\frac{k}{\frac{-1}{3(1+x)^3}} \vert_0^{\infty} \to \frac{-3k}{(1+x)^3} \vert_0^{\infty} = 0 -3k$

and $k= \frac{-1}{3}$

somehow though they came up with $\frac{-k}{3} \frac{1}{(1+x)^3}$ after integrating $\int_0^{\infty} \frac{k}{(1+x)^3}$

Was my approach correct to use substitution? Where does my work go wrong?

Answer 1

The antiderivative of $\frac{1}{u^4}$ is $-\frac13u^{-3}$, not $-3u^{-3}$ as you put. You may check this by differentiating.

Answer 2

Besides the fact that you wrote the antiderivative still under the integral sign, you didn't do the rule right. The antiderivative of $x^r$ is $\frac{x^{r+1}}{r+1}$. So $$ \int_0^\infty \frac1{(1+x)^4}\,dx=\left.\frac{(1+x)^{-3}}{-3}\right|_0^\infty=-\left.\frac{1}{3(1+x)^{3}}\right|_0^\infty=\frac13. $$