Q.Integrate the function $\frac{1}{x-x^3}$

Concept & Intuition

The integral involves a rational function with a repeated linear factor in the denominator. The standard technique is partial fraction decomposition, which rewrites the complicated fraction as a sum of simpler fractions that are easy to integrate. The given answer form already suggests the result will involve a log combination and a single term with $(x-2)^{-1}$. Our job is to find the constants $a$ and $b$ by performing the decomposition and then comparing coefficients.

Step‑by‑Step Solution

1. Set up the partial fraction decomposition Since the denominator is $(x-1)(x-2)^2$, we write:

where $A, B, C$ are constants to be determined.

2. Clear denominators Multiply both sides by $(x-1)(x-2)^2$:

3. Solve for the constants
   • For $C$: Substitute $x = 2$ (makes the $A$ and $B$ terms vanish):

• For $A$: Substitute $x = 1$:

• For $B$: Substitute any convenient value, say $x = 0$, using $A=1, C=2$:

So we have:

4. Integrate term by term

Each integral is elementary:

5. Match with the given form The problem states the result is:

Comparing, we see:

The key insight is that the numerator $x$ is almost the derivative of $x^2$, which appears in the denominator. This suggests a substitution that reduces the quartic denominator to a quadratic in a new variable, making partial fractions straightforward.

1. Substitute $u = x^2$ Let $u = x^2$. Then $du = 2x\,dx$, so $x\,dx = \frac{du}{2}$. The integral becomes

2. Factor the denominator

   Notice $u^2 - 16 = (u-4)(u+4)$. This is a classic setup for partial fractions.

3. Partial fractions decomposition

   We write

Solving: Multiply through by $(u-4)(u+4)$:

Setting $u = 4$ gives $1 = 8A \Rightarrow A = \frac18$.

Setting $u = -4$ gives $1 = -8B \Rightarrow B = -\frac18$.

Hence

4. Integrate

Using logarithm properties:

5. Back-substitute $u = x^2$

Concept & Intuition

The integral is a rational function whose denominator factors into a linear term $(x+2)$ and an irreducible quadratic $(x^2+1)$. The standard method is partial fraction decomposition: we write the integrand as a sum of simpler fractions whose integrals are elementary (logarithms and an arctangent). By comparing the result with the given expression, we can read off the constants $p, q, r$ and then compute their sum.

1. Set up the partial fractions Since the denominator has a linear factor and an irreducible quadratic, we write

The numerator for the quadratic term is linear because the denominator is degree 2.

2. Clear denominators Multiply both sides by $(x+2)(x^2+1)$:

3. Expand and collect like terms

4. Equate coefficients Comparing with the left side $1 = 0x^2 + 0x + 1$ gives the system

5. Solve the system

   From the first equation, $B = -A$.

   Substitute into the second: $2(-A) + C = 0 \Rightarrow C = 2A$.

   Substitute into the third: $A + 2(2A) = 5A = 1 \Rightarrow A = \frac{1}{5}$.

   Then $B = -\frac{1}{5}$ and $C = \frac{2}{5}$.

6. Write the decomposed integrand

7. Integrate term by term

• First integral: $\displaystyle \int\frac{dx}{x+2} = \log|x+2|$.
• Second integral: let $u = x^2+1$, $du = 2x\,dx$ so $\int\frac{x}{x^2+1}dx = \frac{1}{2}\log|x^2+1|$.
• Third integral: $\displaystyle \int\frac{dx}{x^2+1} = \tan^{-1}x$.

Hence

8. Match with the given form The problem states the integral equals

Comparing, we have

9. Compute $p+q+r$

Step 1 — Spot the substitution

Everything inside the bracket is a function of $\log x$, and the leftover $\dfrac{dx}{x}$ is precisely the differential of $\log x$. That is the signal to put

Step 2 — Factorise the quadratic

Split the middle term: $6t^2 + 7t + 2 = 6t^2 + 4t + 3t + 2 = 2t(3t+2) + 1(3t+2)$

Step 3 — Partial fractions

Put $t = -\tfrac12$: $\;1 = B\left(-\tfrac32 + 2\right) = \tfrac{B}{2} \Rightarrow B = 2$.

Put $t = -\tfrac23$: $\;1 = A\left(-\tfrac43 + 1\right) = -\tfrac{A}{3} \Rightarrow A = -3$.

Step 4 — Integrate

Using $\int \frac{dt}{at+b} = \frac{1}{a}\log|at+b|$:

Step 5 — Back-substitute $t = \log x$

Note there is no factor of $\tfrac12$ — the $\tfrac12$ from $\int\frac{dt}{2t+1}$ is cancelled by the numerator $B=2$. That kills options (A) and (D).

ANSWER: B

Solve the integral.

Let $u=\sin x \Rightarrow du=\cos x\,dx$:

Partial fractions:

• At $u=-1:\ -2 = A(1) \Rightarrow A=-2.$
• At $u=-2:\ -4 = B(-1) \Rightarrow B=4.$

So

Comparing with $a\log|1+\sin x| - b\log|2+\sin x| + c$: