# STEP I 2001 5

Recall that:

\displaystyle \int (ax + b)^n \mathrm dx = \frac {(ax + b)^{n + 1}} {a(n + 1)}

for a \ne 0.

(i)

We have:

\begin{align*}\int_0^1 \frac 1 {(1 + tx)^2} \mathrm dx & = \left[-\frac 1 t \times \frac 1 {1 + tx}\right]_0^1 \\ & = -\frac 1 t \times \frac 1 {1 + t} + \frac 1 t \\ & = \frac 1 t - \frac 1 {t(t + 1)} \\ & = \frac t {t(t + 1)} \\ & = \frac 1 {t + 1}\end{align*}

(ii)

Write:

\begin{align*}\int_0^1 \frac {-2x} {(1 + tx)^3} \mathrm dx & = -2 \int_0^1 \frac x {(1 + tx)^3} \mathrm dx \\ & = -\frac 2 t \int_0^1 \frac {tx} {(1 + tx)^3} \mathrm dx \\ & = -\frac 2 t \int_0^1 \frac {tx + 1 - 1} {(1 + tx)^3} \mathrm dx \\ & = -\frac 2 t \int_0^1 \frac 1 {(1 + tx)^2} \mathrm dx + \frac 2 t \int_0^1 \frac 1 {(1 + tx)^3} \mathrm dx \\ & = -\frac 2 t \times \frac 1 {1 + t} - \frac 1 {t^2} \left[\frac 1 {(1 + tx)^2}\right]_0^1 \\ & = -\frac 2 {t(1 + t)} - \frac 1 {t^2 (t + 1)^2} + \frac 1 {t^2} \\ & = \frac {(t + 1)^2 - 1 - 2t(t + 1)} {t^2 (1+t)^2} \\ & = \frac {t^2 + 2t + 1 - 1 - 2t^2 - 2t} {t^2 (1+t)^2} \\ & = -\frac {t^2} {t^2 (1 + t)^2} \\ & = -\frac 1 {(1 + t)^2}\end{align*}

substituting of u = 1 + tx also works.

Note that the derivative of \displaystyle \frac 1 {(1 + tx)^2} with respect to t is \displaystyle \frac {-2x} {(1 + tx)^3}, and the derivative of \displaystyle \int_0^1 \frac 1 {(1 + tx)^2} \mathrm dx is pointed out to be \displaystyle \int_0^1 \frac {-2x} {(1 + tx)^3} \mathrm dx. Since the derivative of \dfrac {-2x} {(1 + tx)^3} with respect to t is \dfrac {6x^2} {(1 + tx)^4}, we should have:

\displaystyle \int_0^1 \frac {6x^2} {(1 + tx)^4} \mathrm dx = \frac {\mathrm d} {\mathrm dt} \left(-\frac 1 {(1 + t)^2}\right) = \frac 2 {(1 + t)^3}

So we conjecture:

\displaystyle \int_0^1 \frac {6x^2} {(1 + x)^4} \mathrm dx = \frac 2 {(1 + 1)^3} = \frac 1 4

setting t = 1.

Optional verification: We “know” that:

\displaystyle \frac {\mathrm d} {\mathrm dt} \left(\int_0^1 \frac {-2x} {(1 + tx)^3} \mathrm dx\right) = \int_0^1 \frac {\partial} {\partial t} \left(\frac {-2x} {(1 + tx)^3}\right) \mathrm dx = \int_0^1 \frac {6x^2} {(1 + x)^4} \mathrm dx

from the Leibniz integral rule: https://en.wikipedia.org/wiki/Leibniz_integral_rule.

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