Q.Evaluate the following definite integrals:

Step 1. (i) Partial fractions. $\dfrac1{x^2-4}=\dfrac1{(x-2)(x+2)}=\dfrac14\left(\dfrac1{x-2}-\dfrac1{x+2}\right)$, so $\displaystyle\int\frac{dx}{x^2-4}=\frac14\ln\left|\frac{x-2}{x+2}\right|+C$.

Step 2. (i) Evaluate at the limits. At $x=4$: $\left|\frac{4-2}{4+2}\right|=\frac13$. At $x=3$: $\left|\frac{3-2}{3+2}\right|=\frac15$.

(This is positive, matching that $1/(x^2-4)>0$ throughout $[3,4]$.)

Step 3. (ii) Complete the square. $x^2+2x+5=(x+1)^2+4=(x+1)^2+2^2$, so $\displaystyle\int\frac{dx}{x^2+2x+5}=\frac12\tan^{-1}\!\left(\frac{x+1}2\right)+C$.

Step 4. (ii) Evaluate at the limits. At $x=1$: $\tan^{-1}(1)=\frac\pi4$. At $x=-1$: $\tan^{-1}(0)=0$.

Step 5. (iii) Rationalize the integrand. Multiply numerator and denominator inside the root by $(1-x)$ (valid since $1-x\geq0$ on $[0,1]$):

Step 6. (iii) Integrate each piece and evaluate. $\displaystyle\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x$, and substituting $u=1-x^2$ gives $\displaystyle\int\frac{x\,dx}{\sqrt{1-x^2}}=-\sqrt{1-x^2}$, so

Step 7. (iv) Rewrite the integrand using half-angle identities. $1+\cos x=2\cos^2\frac x2$ and $\sin x=2\sin\frac x2\cos\frac x2$, so

Step 8. (iv) Recognize the $e^x[f(x)+f'(x)]$ form. With $f(x)=\tan\frac x2$, $f'(x)=\frac12\sec^2\frac x2$, the integrand is exactly $e^x[f(x)+f'(x)]$, whose antiderivative is $e^xf(x)$:

Step 9. (v) Substitute $u=\cos\theta$. Then $du=-\sin\theta\,d\theta$ and $\sin^2\theta=1-u^2$; the limits $\theta=0\to u=1$ and $\theta=\pi/2\to u=0$, and the two sign flips (from $-du$ and from reversing the limits) cancel:

Step 10. (v) Expand and integrate.

Step 11. (vi) Guess and verify the antiderivative by differentiating back. Try $g(x)=\dfrac{x}{1+x^2}$:

which is EXACTLY the given integrand, so $g(x)=\dfrac{x}{1+x^2}$ is a valid antiderivative.

Step 12. (vi) Evaluate at the limits.

Definite integration via partial fractions, completing the square, algebraic/trig substitution, and the e^x[f+f'] rule

• Dropping a sign when reversing the limits of integration after a substitution such as u = cosθ
• Missing the negative sign in ∫x/√(1−x²)dx = −√(1−x²), which flips the sign of that piece in part (iii)
• Not recognizing the e^x[f(x)+f'(x)] pattern in part (iv) and instead attempting integration by parts twice
• In part (vi), reaching for partial fractions or a trig substitution instead of noticing the derivative of x/(1+x²) already matches the integrand exactly