Integration by Substitution: Sure Questions (Class 12, Part 1)

This lesson solves eight Class 12 integrals using the substitution method. In each one we look for an expression whose derivative also appears in the integrand, set that expression equal to $t$, and rewrite the whole integral in terms of $t$.

Powers of a logarithm: $\int \dfrac{(\log x)^2}{x}\,dx$

Notice that $\tfrac{1}{x}$ is the derivative of $\log x$. So put

$t = \log x, \qquad dt = \tfrac{1}{x}\,dx.$

The integral becomes

$\int (\log x)^2 \cdot \tfrac{1}{x}\,dx = \int t^2\,dt = \tfrac{t^3}{3} + c.$

Resubstituting $t = \log x$ gives

$\frac{(\log x)^3}{3} + c.$

A square root with its derivative present: $\int x\sqrt{1+x^2}\,dx$

Here $\tfrac{d}{dx}(1+x^2) = 2x$, and an $x$ already sits in the integrand. Put

$t = 1 + x^2, \qquad dt = 2x\,dx, \qquad x\,dx = \tfrac{1}{2}\,dt.$

Then

$\int \sqrt{1+x^2}\,(x\,dx) = \int \sqrt{t}\cdot\tfrac{1}{2}\,dt = \tfrac{1}{2}\cdot\tfrac{t^{3/2}}{3/2} = \tfrac{1}{3}\,t^{3/2}.$

Resubstituting,

$\frac{1}{3}\,(1+x^2)^{3/2} + c.$

Pulling out a constant first: $\int (4x+2)\sqrt{x^2+x+1}\,dx$

Take $2$ out of $4x+2$ to expose the derivative of $x^2+x+1$, which is $2x+1$:

$\int 2\,(2x+1)\sqrt{x^2+x+1}\,dx.$

Put $t = x^2+x+1$, so $dt = (2x+1)\,dx$. Then

$2\int \sqrt{t}\,dt = 2\cdot\tfrac{t^{3/2}}{3/2} = \tfrac{4}{3}\,t^{3/2},$

and resubstituting gives

$\frac{4}{3}\,(x^2+x+1)^{3/2} + c.$

Deciding to substitute: $\int \dfrac{1}{x-\sqrt{x}}\,dx$

This cannot be integrated directly and will not split, so substitution is the way. Factor $\sqrt{x}$ out of the denominator:

$\frac{1}{x-\sqrt{x}} = \frac{1}{\sqrt{x}\,(\sqrt{x}-1)}.$

Put $t = \sqrt{x}-1$, so $dt = \tfrac{1}{2\sqrt{x}}\,dx$, which means $\tfrac{1}{\sqrt{x}}\,dx = 2\,dt$. The integral becomes

$\int \frac{2\,dt}{t} = 2\log|t| = 2\log|\sqrt{x}-1| + c.$

A logarithm raised to a power: $\int \dfrac{1}{x\,(\log x)^{m}}\,dx$

Again $\tfrac{1}{x}$ is the derivative of $\log x$. Put $t = \log x$, $dt = \tfrac{1}{x}\,dx$:

$\int \frac{1}{t^{m}}\,dt = \int t^{-m}\,dt = \frac{t^{-m+1}}{-m+1}.$

Resubstituting,

$\frac{(\log x)^{1-m}}{1-m} + c.$

An exponential in the denominator: $\int \dfrac{x}{e^{x^2}}\,dx$

Write this as $\int x\,e^{-x^2}\,dx$ and put $t = x^2$, so $dt = 2x\,dx$ and $x\,dx = \tfrac{1}{2}\,dt$:

$\tfrac{1}{2}\int e^{-t}\,dt = \tfrac{1}{2}\,(-e^{-t}) = -\tfrac{1}{2}\,e^{-x^2} + c.$

Ratio of exponentials: $\int \dfrac{e^{2x}-1}{e^{2x}+1}\,dx$

This is not directly integrable, so first divide numerator and denominator by $e^{x}$:

$\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx.$

Now put $t = e^{x}+e^{-x}$, so $dt = (e^{x}-e^{-x})\,dx$, exactly the numerator. Then

$\int \frac{dt}{t} = \log|t| = \log\left|e^{x}+e^{-x}\right| + c.$

A similar exponential ratio: $\int \dfrac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx$

Put $t = e^{2x}+e^{-2x}$. Differentiating,

$dt = \left(2e^{2x}-2e^{-2x}\right)dx = 2\left(e^{2x}-e^{-2x}\right)dx,$

so $(e^{2x}-e^{-2x})\,dx = \tfrac{1}{2}\,dt$. The numerator matches, giving

$\tfrac{1}{2}\int \frac{dt}{t} = \tfrac{1}{2}\log|t| = \tfrac{1}{2}\log\left|e^{2x}+e^{-2x}\right| + c.$

Key takeaways

• Choose a substitution $t$ whose derivative already appears in the integrand, so $dt$ replaces that part cleanly.
• Constants and simple factors (like pulling $2$ out, or factoring $\sqrt{x}$) can be adjusted to reveal the needed derivative.
• For a ratio of exponentials, dividing top and bottom by $e^{x}$ often turns the denominator into something whose derivative is the numerator.