Q.Integrate the following function: $\frac{1}{\sqrt{7 - 6x - x^2}}$

Concept and Intuition

The expression under the square root, $2ax - x^2$, is a quadratic in $x$ that is not a perfect square. The standard trick for integrating $\sqrt{\text{quadratic}}$ is to complete the square so it becomes something like $\sqrt{a^2 - (x - h)^2}$ or $\sqrt{(x - h)^2 + k^2}$. Once we have the form $\sqrt{a^2 - u^2}$, the substitution $u = a\sin\theta$ (or $u = a\cos\theta$) turns the integral into a simple trigonometric one. The final answer will involve an inverse sine (or cosine) and a term like $\frac{u}{2}\sqrt{a^2 - u^2}$.

Step-by-step solution

1. Complete the square

So the integral becomes

2. Choose a substitution Let $u = x - a$, so $du = dx$. Then

This is a classic form. The natural substitution is $u = a\sin\theta$, with $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ (so that $\cos\theta \ge 0$). Then $du = a\cos\theta\, d\theta$, and

3. Transform the integral

4. Integrate $\cos^2\theta$ Use the identity $\cos^2\theta = \frac{1+\cos 2\theta}{2}$:

5. Back-substitute in terms of $u$ Since $\sin 2\theta = 2\sin\theta\cos\theta$, we have

Now $\sin\theta = \frac{u}{a}$ and $\cos\theta = \frac{\sqrt{a^2 - u^2}}{a}$ (positive because of our range). So

Also $\theta = \sin^{-1}\!\left(\frac{u}{a}\right)$.

6. Return to $x$ Recall $u = x - a$, so $$$$

Concept & Intuition

When you see a quadratic inside a square root in an integral, your first instinct should be: complete the square. The expression $2x - x^2$ is a quadratic that opens downward. Rewriting it as $1 - (x-1)^2$ reveals the classic form $\frac{1}{\sqrt{a^2 - u^2}}$, whose antiderivative is $\sin^{-1}(u/a)$. The shift $x-1$ is the key.

Step-by-step solution

1. Complete the square Start with $2x - x^2$. Factor out a negative:

So the integral becomes

2. Recognize the standard form The integrand is now $\frac{1}{\sqrt{1 - u^2}}$ with $u = x-1$. The derivative of $\sin^{-1} u$ is $\frac{1}{\sqrt{1-u^2}}$, so

3. Substitute back Since $u = x-1$, we get

4. Match with options …

Concept: Standard integral - verify by differentiating the candidates.

Claim: d/dx [ -(2/a) * sqrt((a - x)/x) ] = 1 / (x sqrt(ax - x^2))

Let u = (a - x)/x = a/x - 1, so du/dx = -a/x^2.

d/dx [ sqrt(u) ] = (1/(2 sqrt u)) * (-a/x^2)

d/dx [ -(2/a) sqrt(u) ] = -(2/a) * ( -a / (2 x^2 sqrt u) ) = 1 / (x^2 sqrt u)

Now x^2 sqrt u = x^2 * sqrt((a-x)/x) = x^2 * sqrt(a-x)/sqrt(x) = x^(3/2) sqrt(a - x) = x * sqrt(x(a-x)) = x sqrt(ax - x^2). …

We are asked to find $\varphi(x)$ such that $\int \frac{1}{\sqrt{9+8x-x^2}}\,dx = \varphi(x) + c$.

Concept & Intuition

The integrand has a square root of a quadratic that is not a perfect square. The standard technique is to rewrite the quadratic inside the square root by completing the square, so that it resembles $a^2 - (x-h)^2$ or $(x-h)^2 + a^2$. Here, the negative $x^2$ term suggests a downward-opening parabola, so completing the square will give a difference-of-squares form. That form matches the derivative of $\sin^{-1}$ or $\cos^{-1}$, because $\frac{d}{dx}\sin^{-1}\left(\frac{x-h}{a}\right) = \frac{1}{\sqrt{a^2 - (x-h)^2}}$.

Step-by-step solution

1. Complete the square inside the radical. Start with $9 + 8x - x^2$. Factor out a negative from the $x$-terms:

Complete the square for $x^2 - 8x$: half of $-8$ is $-4$, square gives $16$, so

Substitute back:

So the integral becomes

2. Recognize the standard form. The expression $\sqrt{25 - (x-4)^2}$ matches $\sqrt{a^2 - u^2}$ with $a = 5$ and $u = x-4$. The known antiderivative is

3. Apply the formula. Here $du = dx$, so $$$$

Solution

1. Complete the square: $x^2 - 4x + 2 = (x-2)^2 - 2$, so

2. Put $u = x-2,\ du = dx$, with $a^2 = 2$:

3. Apply the standard result

With $a^2 = 2$ the coefficient $\tfrac{a^2}{2} = 1$:

4. Back-substitute $u = x-2$ and $(x-2)^2 - 2 = x^2 - 4x + 2$: …

Write $\csc x-1=\dfrac{1-\sin x}{\sin x}$ and multiply inside the root by $\dfrac{1+\sin x}{1+\sin x}$:

Let $s=\sin x$, $ds=\cos x\,dx$: …

Write $x+2=\tfrac12(2x+2)+1$. Then

The first integral is $\sqrt{x^2+2x+3}$. For the second, $x^2+2x+3=(x+1)^2+2$, so it equals $\log\big|(x+1)+\sqrt{x^2+2x+3}\big|$. Adding: …

Concept & Intuition

When we see $\sqrt{4x^2 - 9}$, it resembles $\sqrt{a^2 x^2 - b^2}$, which suggests a secant substitution. The idea: let $x = \frac{3}{2}\sec\theta$ so that $4x^2 - 9 = 9\sec^2\theta - 9 = 9\tan^2\theta$, and the square root simplifies to $3|\tan\theta|$. This turns the integral into a simple trigonometric form. The key is recognizing that the denominator $x\sqrt{4x^2-9}$ becomes $\frac{3}{2}\sec\theta \cdot 3\tan\theta$, and $dx = \frac{3}{2}\sec\theta\tan\theta\,d\theta$, so the whole thing collapses nicely.

Step-by-step solution

1. Set up the substitution Let $x = \frac{3}{2}\sec\theta$, where $\theta \in [0, \pi/2) \cup (\pi/2, \pi]$ (we take the principal branch where $\tan\theta \ge 0$ for $x \ge 3/2$). Then

2. Simplify the square root

Hence $\sqrt{4x^2-9} = 3|\tan\theta|$. For $x \ge 3/2$, $\theta \in [0, \pi/2)$, so $\tan\theta \ge 0$ and we can drop the absolute value: $\sqrt{4x^2-9} = 3\tan\theta$.

3. Rewrite the integral

4. Integrate

5. Back-substitute Since $x = \frac{3}{2}\sec\theta$, we have $\sec\theta = \frac{2x}{3}$, so $\theta = \sec^{-1}\left(\frac{2x}{3}\right)$. But we can express this in terms of $\tan^{-1}$: …

The key insight here is that the integrand contains a composition of functions that suggests a substitution. Notice the argument of sine is $\tan^{-1}(x^4)$, and the denominator has $1 + x^8$. The derivative of $\tan^{-1}(x^4)$ is $\frac{4x^3}{1 + x^8}$, which appears almost exactly in the numerator — we have $x^3$ and $1 + x^8$, just missing the factor of 4. This is a classic setup for a substitution that turns the integral into something like $\int \sin u \, du$.

Let’s work through it step by step.

1. Choose the substitution. Let $u = \tan^{-1}(x^4)$. Then differentiate:

Rearranging gives:

2. Rewrite the integral in terms of $u$. The original integral is:

We have $du = \frac{4x^3}{1 + x^8} dx$, so $\frac{x^3}{1 + x^8} dx = \frac{du}{4}$.

Also, $\sin(\tan^{-1}(x^4)) = \sin u$.

Therefore:

3. Integrate. The integral of $\sin u$ is $-\cos u$, so:

4. Substitute back. Since $u = \tan^{-1}(x^4)$, we get: $I = -\frac{\cos(\tan^{-1}(x^4))}{4} + C.$ …

Write

so that $2x-1=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)$.

Using the cover-up method (substitute each root):

• $x=1:\quad 2(1)-1=A(3)(-2)\ \Rightarrow\ 1=-6A\ \Rightarrow\ A=-\dfrac{1}{6}.$
• $x=-2:\quad 2(-2)-1=B(-3)(-5)\ \Rightarrow\ -5=15B\ \Rightarrow\ B=-\dfrac{1}{3}.$ …