Integration: Miscellaneous Exercise 7 Solutions (Part 3)

This lesson works through three integrals from the Class 12 Miscellaneous Exercise on integration: one needing a substitution with integration by parts, one general power rule, and one partial-fractions problem.

Question 23

Evaluate

$I = \int \frac{\sqrt{x^2+1}\,\bigl(\log(x^2+1) - 2\log x\bigr)}{x^4}\,dx.$

Combine the logarithms. Using $2\log x = \log x^2$ and $\log a - \log b = \log\tfrac{a}{b}$,

$\log(x^2+1) - 2\log x = \log\frac{x^2+1}{x^2}.$

Pull the root inside. Since $\sqrt{x^2+1} = x\sqrt{\dfrac{x^2+1}{x^2}}$, and writing $x^4 = x\cdot x^3$,

$I = \int \sqrt{\frac{x^2+1}{x^2}}\;\log\!\frac{x^2+1}{x^2}\;\frac{1}{x^3}\,dx.$

Substitute. Let

$t = \frac{x^2+1}{x^2} = 1 + \frac{1}{x^2}, \qquad dt = -\frac{2}{x^3}\,dx \;\Rightarrow\; \frac{1}{x^3}\,dx = -\frac{1}{2}\,dt.$

Then

$I = -\frac{1}{2}\int \sqrt{t}\,\log t\,dt.$

Integration by parts with $\log t$ as the first function and $\sqrt{t}$ as the second:

$\int t^{1/2}\log t\,dt = \frac{2}{3}t^{3/2}\log t - \int \frac{1}{t}\cdot\frac{2}{3}t^{3/2}\,dt = \frac{2}{3}t^{3/2}\log t - \frac{4}{9}t^{3/2}.$

Multiplying by $-\tfrac{1}{2}$,

$I = -\frac{1}{3}t^{3/2}\log t + \frac{2}{9}t^{3/2} = -\frac{1}{3}t^{3/2}\left(\log t - \frac{2}{3}\right) + C.$

Back-substitute $t = \dfrac{x^2+1}{x^2}$, using $\left(\dfrac{x^2+1}{x^2}\right)^{3/2} = \dfrac{(x^2+1)^{3/2}}{x^3}$:

$I = -\frac{1}{3}\,\frac{(x^2+1)^{3/2}}{x^3}\left(\log\frac{x^2+1}{x^2} - \frac{2}{3}\right) + C.$

Question 17

Evaluate

$I = \int \bigl(f(ax+b)\bigr)^n\,f'(ax+b)\,dx.$

Substitute $t = f(ax+b)$. Differentiating, the chain rule gives a factor $a$ from $ax+b$:

$dt = a\,f'(ax+b)\,dx \;\Rightarrow\; f'(ax+b)\,dx = \frac{1}{a}\,dt.$

So

$I = \frac{1}{a}\int t^{n}\,dt = \frac{1}{a}\cdot\frac{t^{n+1}}{n+1} + C.$

Back-substituting $t = f(ax+b)$,

$I = \frac{\bigl(f(ax+b)\bigr)^{n+1}}{a\,(n+1)} + C.$

Question 21

Evaluate

$I = \int \frac{x^2+x+1}{(x+1)^2\,(x+2)}\,dx.$

Partial fractions. Because $(x+1)$ is repeated,

$\frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{a}{x+1} + \frac{b}{(x+1)^2} + \frac{c}{x+2}.$

Clearing denominators,

$x^2+x+1 = a(x+1)(x+2) + b(x+2) + c(x+1)^2.$

Find $b$: put $x = -1$: $\;1 = b(1)$, so $b = 1$.

Find $c$: put $x = -2$: $\;3 = c(1)$, so $c = 3$.

Find $a$: compare the $x^2$ coefficients: $1 = a + c = a + 3$, so $a = -2$.

Therefore

$I = \int\left(\frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2}\right)dx.$

Integrating term by term,

$I = -2\log|x+1| - \frac{1}{x+1} + 3\log|x+2| + C.$

Combine the logs using $-2\log|x+1| = -\log(x+1)^2$ and $3\log|x+2| = \log|x+2|^3$:

$I = \log\frac{|x+2|^3}{(x+1)^2} - \frac{1}{x+1} + C.$

Key takeaways

• Combining logarithms and choosing a clean substitution can turn a complicated integrand into a standard $\sqrt{t}\,\log t$ form for integration by parts.
• For $\int (f(ax+b))^n f'(ax+b)\,dx$, substituting $t = f(ax+b)$ gives $\dfrac{(f(ax+b))^{n+1}}{a(n+1)} + C$.
• A repeated linear factor in the denominator needs both a $\dfrac{1}{x+1}$ and a $\dfrac{1}{(x+1)^2}$ term in the partial-fraction split.