Partial Fractions by Substitution (Part 1)

This lesson works through four integrals from the Class 12 sure-question set where a substitution first simplifies the integrand, and partial fractions then finish the job. In each case we choose a substitution, reduce the integral to a simple rational function, split it into partial fractions, integrate, and substitute back.

First integral: $\int \dfrac{2x}{(x^2+1)(x^2+3)}\,dx$

Because the numerator $2x$ is exactly the derivative of $x^2$, substitute $x^2 = t$, so $2x\,dx = dt$. The integral becomes

$I = \int \frac{dt}{(t+1)(t+3)}.$

Partial fractions

Write

$\frac{1}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3},$

so $1 = A(t+3) + B(t+1)$.

• Put $t = -3$: $\;1 = -2B$, hence $B = -\tfrac{1}{2}$.
• Put $t = -1$: $\;1 = 2A$, hence $A = \tfrac{1}{2}$.

Integrate

$I = \tfrac{1}{2}\int \frac{dt}{t+1} - \tfrac{1}{2}\int \frac{dt}{t+3} = \tfrac{1}{2}\ln|t+1| - \tfrac{1}{2}\ln|t+3|.$

Using $\ln a - \ln b = \ln\tfrac{a}{b}$ and resubstituting $t = x^2$,

$I = \tfrac{1}{2}\ln\left|\frac{x^2+1}{x^2+3}\right| + C.$

Second integral: $\int \dfrac{1}{a^{x}-1}\,dx$

Put $a^{x} = t$. Differentiating, $a^{x}\ln a\,dx = dt$, so $dx = \dfrac{dt}{a^{x}\ln a} = \dfrac{dt}{t\ln a}$. Then

$I = \int \frac{1}{t-1}\cdot\frac{dt}{t\ln a} = \frac{1}{\ln a}\int \frac{dt}{t(t-1)}.$

Partial fractions

$\frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1}, \qquad 1 = A(t-1) + Bt.$

• Put $t = 1$: $\;1 = B$, so $B = 1$.
• Put $t = 0$: $\;1 = -A$, so $A = -1$.

Integrate

$I = \frac{1}{\ln a}\left[-\int \frac{dt}{t} + \int \frac{dt}{t-1}\right] = \frac{1}{\ln a}\left(-\ln|t| + \ln|t-1|\right) = \frac{1}{\ln a}\ln\left|\frac{t-1}{t}\right|.$

Resubstituting $t = a^{x}$,

$I = \frac{1}{\ln a}\ln\left|\frac{a^{x}-1}{a^{x}}\right| + C.$

Third integral: $\int \dfrac{x^{3}}{x^{4}+3x^{2}+2}\,dx$

Write the numerator as $x^{2}\cdot x$ and the denominator as $(x^{2})^{2} + 3x^{2} + 2$. Put $x^{2} = t$, so $2x\,dx = dt$, i.e. $x\,dx = \tfrac{1}{2}\,dt$. Then

$I = \frac{1}{2}\int \frac{t\,dt}{t^{2}+3t+2} = \frac{1}{2}\int \frac{t\,dt}{(t+1)(t+2)}.$

Partial fractions

$\frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}, \qquad t = A(t+2) + B(t+1).$

• Put $t = -2$: $\;-2 = -B$, so $B = 2$.
• Put $t = -1$: $\;-1 = A$, so $A = -1$.

Integrate

$I = \frac{1}{2}\left[-\int \frac{dt}{t+1} + 2\int \frac{dt}{t+2}\right] = -\tfrac{1}{2}\ln|t+1| + \ln|t+2|.$

Resubstituting $t = x^{2}$,

$I = \ln|x^{2}+2| - \tfrac{1}{2}\ln|x^{2}+1| + C.$

Fourth integral: $\int \dfrac{x+1}{x\,(1 + x\,e^{x})}\,dx$

The key first step: multiply numerator and denominator by $e^{x}$.

$I = \int \frac{(x+1)\,e^{x}}{x\,e^{x}\,(1 + x\,e^{x})}\,dx.$

Now put $x\,e^{x} = t$. By the product rule, $\dfrac{d}{dx}(x\,e^{x}) = x\,e^{x} + e^{x} = (x+1)\,e^{x}$, so $(x+1)\,e^{x}\,dx = dt$. The numerator and the leftover factor $x\,e^{x}$ in the denominator become $dt$ and $t$, giving

$I = \int \frac{dt}{t\,(1+t)}.$

Partial fractions

$\frac{1}{t(1+t)} = \frac{A}{t} + \frac{B}{1+t}, \qquad 1 = A(1+t) + Bt.$

• Put $t = -1$: $\;1 = -B$, so $B = -1$.
• Put $t = 0$: $\;1 = A$, so $A = 1$.

Integrate

$I = \int \frac{dt}{t} - \int \frac{dt}{1+t} = \ln|t| - \ln|1+t| = \ln\left|\frac{t}{1+t}\right|.$

Resubstituting $t = x\,e^{x}$,

$I = \ln\left|\frac{x\,e^{x}}{1 + x\,e^{x}}\right| + C.$

Key takeaways

• When the numerator is the derivative of a power of $x$, substitute that power equal to $t$ to collapse the integral into a simple rational function.
• After substituting, split into partial fractions and find the constants by putting in convenient values of $t$.
• Each piece integrates to a logarithm; combine them with $\ln a - \ln b = \ln\tfrac{a}{b}$ and remember to substitute the original variable back in.