Integration miscellaneous exercise 7: first five questions

This lesson solves the first five integrals of the Class 12 miscellaneous exercise on integration. Each one needs a different idea: partial fractions, a conjugate, or a substitution that simplifies a surd.

Question 1: $\displaystyle\int \frac{1}{x - x^3}\,dx$

Factor the denominator first:

$x - x^3 = x(1 - x^2) = x(1 + x)(1 - x).$

So we write the integrand in partial fractions:

$\frac{1}{x(1 + x)(1 - x)} = \frac{A}{x} + \frac{B}{1 + x} + \frac{C}{1 - x}.$

Multiplying through by $x(1+x)(1-x)$ and substituting handy values of $x$:

Finding the constants

• $x = 0$: $\;1 = A(1)(1) \Rightarrow A = 1.$
• $x = -1$: $\;1 = B(-1)(2) \Rightarrow B = -\tfrac{1}{2}.$
• $x = 1$: $\;1 = C(1)(2) \Rightarrow C = \tfrac{1}{2}.$

Now integrate term by term:

$I = \int \frac{1}{x}\,dx - \tfrac{1}{2}\int \frac{1}{1 + x}\,dx + \tfrac{1}{2}\int \frac{1}{1 - x}\,dx.$

Note $\int \frac{1}{1-x}\,dx = -\ln|1 - x|$, so

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

Write $\ln|x| = \tfrac{1}{2}\ln x^2$ and combine the last two logs:

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

Question 2: $\displaystyle\int \frac{1}{\sqrt{x + a} + \sqrt{x + b}}\,dx$

Multiply numerator and denominator by the conjugate $\sqrt{x + a} - \sqrt{x + b}$:

$\frac{\sqrt{x + a} - \sqrt{x + b}}{(x + a) - (x + b)} = \frac{\sqrt{x + a} - \sqrt{x + b}}{a - b}.$

So

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

Using $\int (x + k)^{1/2}\,dx = \tfrac{2}{3}(x + k)^{3/2}$:

$I = \frac{2}{3(a - b)}\left[(x + a)^{3/2} - (x + b)^{3/2}\right] + C.$

Question 3: $\displaystyle\int \frac{1}{x\sqrt{ax - x^2}}\,dx$

The key step is the substitution $x = \dfrac{a}{t}$, so that $dx = -\dfrac{a}{t^2}\,dt$ and $t = \dfrac{a}{x}$.

The square root becomes

$\sqrt{ax - x^2} = \sqrt{\frac{a^2}{t} - \frac{a^2}{t^2}} = \frac{a}{t}\sqrt{t - 1}.$

Substituting everything in, the factors of $a$ and $t$ cancel and we are left with

$I = -\frac{1}{a}\int \frac{dt}{\sqrt{t - 1}} = -\frac{2}{a}\sqrt{t - 1} + C.$

Putting $t = \dfrac{a}{x}$ back:

$I = -\frac{2}{a}\sqrt{\frac{a - x}{x}} + C.$

Question 4: $\displaystyle\int \frac{1}{x^2\,(x^4 + 1)^{3/4}}\,dx$

Take $x^4$ out of the bracket: $(x^4 + 1)^{3/4} = x^3\,(1 + x^{-4})^{3/4}$. Then

$I = \int \frac{1}{x^5}\,(1 + x^{-4})^{-3/4}\,dx.$

Let $t = 1 + x^{-4}$, so $dt = -4x^{-5}\,dx$, giving $x^{-5}\,dx = -\tfrac{1}{4}\,dt$:

$I = -\frac{1}{4}\int t^{-3/4}\,dt = -\frac{1}{4}\cdot \frac{t^{1/4}}{1/4} = -\,t^{1/4} + C.$

Returning to $x$:

$I = -\left(1 + x^{-4}\right)^{1/4} + C.$

Question 5: $\displaystyle\int \frac{1}{x^{1/2} + x^{1/3}}\,dx$

Let $x = t^6$ (so $x^{1/6} = t$), giving $dx = 6t^5\,dt$. Then $x^{1/2} = t^3$ and $x^{1/3} = t^2$, so the denominator is $t^3 + t^2 = t^2(t + 1)$:

$I = \int \frac{6t^5}{t^2(t + 1)}\,dt = 6\int \frac{t^3}{t + 1}\,dt.$

Long division

Dividing $t^3$ by $t + 1$ gives

$\frac{t^3}{t + 1} = t^2 - t + 1 - \frac{1}{t + 1}.$

Integrating term by term:

$I = 6\left(\frac{t^3}{3} - \frac{t^2}{2} + t - \ln|t + 1|\right) + C.$

With $t = x^{1/6}$:

$I = 2\sqrt{x} - 3\,x^{1/3} + 6\,x^{1/6} - 6\ln\left|x^{1/6} + 1\right| + C.$

Key takeaways

• A rational function with distinct linear factors splits into partial fractions you can integrate one at a time.
• Multiplying by the conjugate clears a sum of square roots from the denominator.
• A substitution like $x = a/t$ or $x = t^6$ can turn an awkward surd into a standard power integral.
• Always rewrite the final answer back in terms of the original variable $x$.