The indefinite integral $\displaystyle\int{\frac{\mathrm{P}(x)}{{{x}^{3}}+1}\,\mathrm{d}x}$, where $\mathrm{P}\left( x \right)$ is a polynomial in $x$, is denoted by $\mathrm{I}$.
<ol class="i-list"><li><div class="list-item-paragraph">Find $\mathrm{I}$ when $\mathrm{P}(x)={{x}^{2}}$.</div>[2]</li><li><div class="list-item-paragraph">By writing ${{x}^{3}}+1=\left( x+1 \right)\left( {{x}^{2}}+Ax+B \right)$, where $A$ and $B$ are constants, find $\mathrm{I}$ when</div><ol class="abc-list"><li><div class="list-item-paragraph">$\mathrm{P}\left( x \right)={{x}^{2}}-x+1$,</div>[3]</li><li><div class="list-item-paragraph">$\mathrm{P}\left( x \right)=x+1$.</div>[3]</li></ol></li><li><div class="list-item-paragraph">Using the results of parts (i) and (ii), or otherwise, find $\mathrm{I}$ when $\mathrm{P}\left( x \right)=1$.</div>[4]</li></ol>

The indefinite integral $\displaystyle\displaystyle \displaystyle\int{\frac{\mathrm{P}(x)}{{{x}^{3}}+1}\,\mathrm{d}x}$, where $\mathrm{P}\left( x \right)$ is a polynomial in $x$, is denoted by $\mathrm{I}$.

Find $\mathrm{I}$ when $\mathrm{P}(x)={{x}^{2}}$.

By writing ${{x}^{3}}+1=\left( x+1 \right)\left( {{x}^{2}}+Ax+B \right)$, where $A$ and $B$ are constants, find $\mathrm{I}$ when

Using the results of parts **(i)** and **(ii)**, or otherwise, find $\mathrm{I}$ when $\mathrm{P}\left( x \right)=1$.

Default solution

$I = \displaystyle\int \dfrac{P(x)}{x^3+1}\,dx$, $P(x) = x^2$\[= \dfrac{1}{3}\int \dfrac{3x^2}{x^3+1}\,dx\]Let $u = x^3 + 1$:\[= \dfrac{1}{3}\ln|x^3 + 1| + C\]

$P(x) = 1$\[I = \int \dfrac{1}{x^3+1}\,dx = \int \dfrac{1}{(x+1)(x^2-x+1)}\,dx\]Partial fractions: $\dfrac{1}{(x+1)(x^2-x+1)} = \dfrac{A}{x+1} + \dfrac{Bx+C}{x^2-x+1}$$1 \equiv A(x^2-x+1) + (Bx+C)(x+1)$$x = -1$: $1 = A(3) \implies A = \dfrac{1}{3}$$x = 0$: $1 = \dfrac{1}{3}(1) + C(1) \implies C = \dfrac{2}{3}$Coeff of $x$: $0 = A + B \implies B = -\dfrac{1}{3}$\[= \int \dfrac{1}{3(x+1)} + \dfrac{-\frac{1}{3}x + \frac{2}{3}}{x^2-x+1}\,dx\]\[= \dfrac{1}{3}\ln|x+1| - \dfrac{1}{3}\int \dfrac{x - 2}{x^2-x+1}\,dx\]Write $x - 2 = \dfrac{1}{2}(2x-1) - \dfrac{3}{2}$:\[= \dfrac{1}{3}\ln|x+1| - \dfrac{1}{6}\int \dfrac{2x-1}{x^2-x+1}\,dx + \dfrac{1}{2}\int \dfrac{1}{x^2-x+1}\,dx\]\[= \dfrac{1}{3}\ln|x+1| - \dfrac{1}{6}\ln|x^2-x+1| + \dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x-1}{\sqrt{3}}\right) + C\]

Primary Solution

Integration by Splitting the Numerator

Techniques of Integration

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

IB AA HL Math

Integration

N2005 P1 Q14 [Either]

$P(x) = x^2 - x + 1$Since $x^3 + 1 = (x+1)(x^2 - x + 1)$:\[I = \int \dfrac{x^2 - x + 1}{(x+1)(x^2-x+1)}\,dx = \int \dfrac{1}{x+1}\,dx\]\[= \ln|x+1| + C\]

$P(x) = x + 1$\[I = \int \dfrac{x+1}{(x+1)(x^2-x+1)}\,dx = \int \dfrac{1}{x^2 - x + 1}\,dx\]Complete the square: $x^2 - x + 1 = \left(x - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}$\[= \int \dfrac{1}{\left(x - \dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2}\,dx\]\[= \dfrac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\left(\dfrac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C = \dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x-1}{\sqrt{3}}\right) + C\]

N2005 P1 Q14 [Either]

N2005 P1 Q14 [Either]

Video Solution:

Default solution