Q.Evaluate: $\int \dfrac{\sqrt{1+x^2}}{x^4}\,dx$

Concept and Intuition

Integrands that mix $x$, $1+x$, and $\sqrt{x(1+x)}$ often simplify dramatically if you introduce $a=\sqrt x$ and $b=\sqrt{1+x}$, because $a^2+ab$ or $b^2+ab$ factor as $a(a+b)$ or $b(a+b)$ — exactly cancelling a denominator of $a+b=\sqrt x+\sqrt{1+x}$.

Step-by-Step Solution

1. Let $a=\sqrt x$, $b=\sqrt{1+x}$. Then $a^2=x$, $b^2=1+x$, and $\sqrt{x+x^2}=\sqrt{x(1+x)}=\sqrt{a^2b^2}=ab$ (for $x\ge0$).
2. Numerator: $1+x+\sqrt{x+x^2} = b^2 + ab = b(b+a)$.
3. Denominator: $\sqrt x + \sqrt{1+x} = a+b$.
4. The integrand is $\dfrac{b(a+b)}{a+b} = b = \sqrt{1+x}$ (for $x>-1$, $a+b\ne 0$).
5. So the integral reduces to $\displaystyle\int \sqrt{1+x}\,dx$. …

Concept and Intuition

Before integrating a rational function, always check whether the numerator has the denominator as a factor — grouping terms often reveals this cleanly.

Step-by-Step Solution

1. Group: $x^3-x^2+x-1=x^2(x-1)+1\cdot(x-1)=(x-1)(x^2+1)$.
2. So $\dfrac{x^3-x^2+x-1}{x-1}=x^2+1$ (for $x\ne1$).
3. $\displaystyle\int(x^2+1)\,dx=\dfrac{x^3}{3}+x+c$.

Common Mistakes …

Concept and Intuition

Integrals mixing $\cos x$, $\sin x$ and $\sin 2x$ often simplify beautifully once everything is expressed in terms of $\tan x$, because $\sec^2x\,dx = d(\tan x)$ absorbs the $dx$ cleanly.

Step-by-Step Solution

1. $\sin 2x = 2\sin x\cos x$, so $\sqrt{2\sin 2x} = \sqrt{4\sin x\cos x} = 2\sqrt{\sin x\cos x}$.
2. The integrand is $\dfrac{1}{\cos^3x\cdot 2\sqrt{\sin x\cos x}} = \dfrac{1}{2\cos^{7/2}x\,\sin^{1/2}x}$.
3. Write $\cos^{7/2}x\sin^{1/2}x = \cos^4x\cdot\left(\dfrac{\sin x}{\cos x}\right)^{1/2} = \cos^4x\,(\tan x)^{1/2}$.
4. So the integrand is $\dfrac{\sec^4x}{2(\tan x)^{1/2}}dx$.
5. Let $t=\tan x$, $dt=\sec^2x\,dx$. Then $\sec^4x\,dx = \sec^2x\cdot\sec^2x\,dx = (1+t^2)dt$. …

Concept and Intuition

When an integrand mixes $\sin x$ and $\cos x$ with half-integer/odd powers overall matching a $\sec^2x\,dx=dt$ substitution, converting fully to $\tan x=t$ collapses the trigonometric mess into ordinary powers of $t$.

Step-by-Step Solution

1. $\sqrt{2\sin2x}=\sqrt{4\sin x\cos x}=2\sqrt{\sin x\cos x}$, so the integrand is

2. Let $t=\tan x$, $dt=\sec^2x\,dx$, so $dx=\cos^2x\,dt$. Write $\cos^{-7/2}x=\sec^2x\cdot\cos^{-3/2}x$, absorbing the $\sec^2x$ into $dt$:

3. Using $\sin x=t\cos x$: $\sin^{-1/2}x=t^{-1/2}\cos^{-1/2}x$, so $\sin^{-1/2}x\cos^{-3/2}x=t^{-1/2}\cos^{-2}x=t^{-1/2}(1+t^2)$ (since $\sec^2x=1+t^2$). …

Concept and Intuition

Dividing termwise turns the awkward-looking fraction into a sum of a monomial and a simple power of $x$, both of which integrate elementarily.

Step-by-Step Solution

1. $\dfrac{x^3-1}{x^2} = \dfrac{x^3}{x^2}-\dfrac{1}{x^2} = x-x^{-2}$.
2. Antiderivative: $\displaystyle\int(x-x^{-2})\,dx = \dfrac{x^2}{2}+\dfrac1x+C$.
3. At $x=2$: $\dfrac{4}{2}+\dfrac12 = 2+0.5=2.5$. …

Concept and Intuition

A definite integral with a variable limit is just the antiderivative evaluated at the two limits — no need for anything fancier here since the integrand is a simple polynomial.

Step-by-Step Solution

1. Antiderivative of $1-t$ is $t-\dfrac{t^2}{2}$.
2. $\displaystyle\int_x^1(1-t)dt = \left[t-\frac{t^2}{2}\right]_x^1 = \left(1-\frac12\right)-\left(x-\frac{x^2}{2}\right)=\frac12-x+\frac{x^2}{2}$.
3. Set equal to $\dfrac12$: $\dfrac12-x+\dfrac{x^2}{2}=\dfrac12\Rightarrow \dfrac{x^2}{2}-x=0\Rightarrow x\left(\dfrac{x}{2}-1\right)=0$.
4. So $x=0$ or $x=2$. …