Theoretical Recall

Substitution rule:

If $x=g(t)$ is a differentiable substitution, then

\[\int f(x)\,dx = \int f(g(t)) g'(t)\, dt.\]

Common strategies:

  • Simplify radicals by trigonometric substitution ($x=\sin t$, $x=\tan t$).
  • Linear substitutions to reduce shifts/scaling.
  • Exponential/logarithmic substitutions to handle $\log$, $e^x$, rational forms.

Author’s note: Always transform both the function and the differential $dx$ consistently.

Exercises

Exercise 1

Evaluate \(\int (2x+1)^5 \, dx.\)

Solution:
Let $u=2x+1$, $du=2dx$, $dx=\tfrac{1}{2}du$.

\[\int (2x+1)^5 dx = \tfrac{1}{2}\int u^5 du = \tfrac{1}{12}u^6+C.\]

Back-substitute: $(2x+1)^6/12+C$.

Final Result: \(\int (2x+1)^5 dx = \frac{(2x+1)^6}{12}+C\)

Exercise 2

Evaluate \(\int \frac{1}{x\log x}\, dx, \quad x>1.\)

Solution:
Let $u=\log x$, $du=\tfrac{1}{x}dx$.

\[\int \frac{1}{x\log x} dx = \int \frac{1}{u} du = \log|u|+C=\log(\log x)+C.\]

Final Result: \(\int \frac{1}{x\log x} dx = \log(\log x)+C\)

Exercise 3

Evaluate \(\int \frac{x}{1+x^2}\, dx.\)

Solution:
Let $u=1+x^2$, $du=2x dx$.

\[\int \frac{x}{1+x^2} dx = \tfrac{1}{2}\int \frac{1}{u} du = \tfrac{1}{2}\log(1+x^2)+C.\]

Final Result: \(\int \frac{x}{1+x^2} dx = \tfrac{1}{2}\log(1+x^2)+C\)

Exercise 4

Evaluate \(\int \frac{1}{\sqrt{1-x^2}} dx.\)

Solution:
Substitute $x=\sin t$, $dx=\cos t dt$.

\[\int \frac{1}{\sqrt{1-\sin^2 t}} \cos t dt = \int 1 dt = t+C.\]

Back-substitute: $\arcsin x + C$.

Final Result: \(\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x + C\)

Exercise 5

Evaluate \(\int \frac{1}{x^2+1} dx.\)

Solution:
Let $x=\tan t$, $dx=\sec^2 t dt$, $1+x^2=\sec^2 t$.

\[\int \frac{1}{1+\tan^2 t} \sec^2 t dt = \int 1 dt = t+C.\]

Back-substitute: $\arctan x + C$.

Final Result: \(\int \frac{1}{x^2+1} dx = \arctan x + C\)

Exercise 6

Evaluate \(\int \frac{1}{\sqrt{x^2+4}} dx.\)

Solution:
Substitute $x=2\tan t$, $dx=2\sec^2 t dt$.

\[\sqrt{x^2+4}=\sqrt{4\tan^2 t+4}=2\sec t.\]

So integral:
\(\int \frac{2\sec^2 t}{2\sec t} dt = \int \sec t dt.\)

Result: $\log \sec t+\tan t +C$. Back-substitute $t=\arctan(x/2)$.

Final Result: \(\int \frac{1}{\sqrt{x^2+4}} dx = \log\!\left|\frac{x+\sqrt{x^2+4}}{2}\right|+C\)

Exercise 7

Evaluate \(\int \frac{1}{x^2-1} dx.\)

Solution:
Use partial fraction substitution:
\(\frac{1}{x^2-1}=\tfrac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right).\)

Then integrate directly.

Final Result: \(\int \frac{1}{x^2-1} dx = \tfrac{1}{2}\log\!\left|\frac{x-1}{x+1}\right|+C\)

Exercise 8

Evaluate \(\int \frac{1}{\sqrt{x^2-1}} dx, \quad x>1.\)

Solution:
Set $x=\cosh t$, $dx=\sinh t dt$, $\sqrt{x^2-1}=\sinh t$.

\[\int \frac{1}{\sinh t}\sinh t dt = \int 1 dt = t+C.\]

Back-substitute $t=\operatorname{arcosh}(x)=\log(x+\sqrt{x^2-1})$.

Final Result: \(\int \frac{1}{\sqrt{x^2-1}} dx = \log(x+\sqrt{x^2-1})+C\)

Exercise 9

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

Solution:
Set $x=\sinh t$, $dx=\cosh t dt$.

\[\int \sqrt{1+\sinh^2 t}\cosh t dt = \int \cosh^2 t dt.\]

Use identity $\cosh^2 t=\tfrac{1}{2}(\cosh 2t+1)$.
Integral $=\tfrac{1}{4}\sinh 2t+\tfrac{1}{2}t+C$.
Back-substitute $t=\operatorname{arsinh}(x)$.

Final Result: \(\int \sqrt{1+x^2}\, dx = \tfrac{1}{2}\left(x\sqrt{1+x^2}+\operatorname{arsinh}(x)\right)+C\)

Exercise 10

Evaluate \(\int \frac{1}{x\sqrt{x^2-1}} dx, \quad x>1.\)

Solution:
Substitute $x=\sec t$, $dx=\sec t\tan t dt$.

\[\int \frac{\sec t \tan t}{\sec t \tan t} dt = \int 1 dt = t+C.\]

Back-substitute: $t=\operatorname{arcsec}(x)$.

Final Result: \(\int \frac{1}{x\sqrt{x^2-1}} dx = \operatorname{arcsec}(x)+C\)