NCERT Solutions Class 12 Maths Integrals Exercise 7.4

The Six Standard Special Integral Formulas

• $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C$
• $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log\left|\frac{a+x}{a-x}\right| + C$
• $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a} ight) + C$
• $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + C$
• $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a} ight) + C$
• $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$

Method of Completing the Square

To successfully solve problems in class 12 maths integrals ex 7 4, you must master the art of reducing quadratic expressions to standard forms. When faced with an integrand containing a quadratic expression $ax^2 + bx + c$, your primary goal is to convert it into the form $a[(x + p)^2 \pm q^2]$ by completing the square. First, make the coefficient of $x^2$ unity by factoring out $a$. Next, add and subtract the square of half the coefficient of $x$. This simple algebraic rearrangement shifts the quadratic term into a perfect square, allowing you to substitute $t = x + p$ and directly apply one of the six standard integration formulas. When the quadratic is inside a square root, such as $\sqrt{ax^2+bx+c}$, extra caution must be taken if $a < 0$. You must factor out the negative sign inside the square root but keep it under the root, rewriting the expression as $\sqrt{-a[x^2 + \frac{b}{a}x + \frac{c}{a}]}$ and manipulating the signs carefully to avoid imaginary terms.

Step-by-Step Integration of Linear/Quadratic Forms

1. Express Numerator as Derivative of Denominator — Write the numerator $px+q$ as a linear combination of the derivative of the quadratic expression plus a constant: $px + q = A \frac{d}{dx}(ax^2+bx+c) + B = A(2ax + b) + B$.
2. Determine Constants A and B — Equate the coefficients of $x$ and the constant terms on both sides of the equation to find the exact values of $A$ and $B$.
3. Split the Integral — Rewrite the original integral as two separate terms: $I = A \int \frac{2ax+b}{ax^2+bx+c} dx + B \int \frac{1}{ax^2+bx+c} dx$.
4. Integrate Independently — The first integral is solved easily by substitution ($t = ax^2+bx+c \implies dt = (2ax+b)dx$), resulting in a logarithmic term. The second integral is solved by completing the square and applying the standard special formulas.

Avoid These Common Board Exam Mistakes

One of the most common mistakes students make in integrals ex 7 4 ncert notes is forgetting to divide by the coefficient of $x^2$ when completing the square, or applying the wrong sign formula. For instance, when integrating $\frac{1}{a^2-x^2}$, students frequently write the formula for $\frac{1}{x^2-a^2}$ by mistake. Always double-check your formula signs! Another deadly trap: if you have a negative coefficient inside a square root, such as $\sqrt{5-4x-x^2}$, do NOT pull the negative sign outside the square root as that would create an imaginary unit ($i$). Instead, keep the negative sign inside the bracket: $\sqrt{-(x^2+4x-5)} = \sqrt{9 - (x+2)^2}$.

Practice Questions with Solutions

• Q: Find the integral of $\frac{1}{x^2 - 16}$ with respect to $x$. A: Step 1: Write the integrand in standard form. Here, the denominator is $x^2 - 16 = x^2 - 4^2$. Thus, $a = 4$. Step 2: Identify the correct formula: $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C$. Step 3: Substitute $a = 4$ into the formula: $\int \frac{dx}{x^2 - 4^2} = \frac{1}{2(4)} \log\left|\frac{x-4}{x+4}\right| + C$. Final answer: $\frac{1}{8} \log\left|\frac{x-4}{x+4}\right| + C$
• Q: Evaluate $\int \frac{dx}{x^2 + 2x + 5}$. A: Step 1: Complete the square for the denominator: $x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2$. Step 2: Let $t = x+1$, which gives $dt = dx$. The integral becomes $\int \frac{dt}{t^2 + 2^2}$. Step 3: Apply the standard formula $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C$ where $a = 2$. Step 4: Substitute back $t = x+1$ to get the final solution: $\frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C$. Final answer: $\frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C$
• Q: Evaluate the integral $\int \frac{dx}{\sqrt{9 - 25x^2}}$. A: Step 1: Standardise the quadratic term by factoring out $25$ from the terms inside the root: $\sqrt{9 - 25x^2} = \sqrt{25\left(\frac{9}{25} - x^2\right)} = 5\sqrt{\left(\frac{3}{5}\right)^2 - x^2}$. Step 2: Rewrite the integral as $\frac{1}{5} \int \frac{dx}{\sqrt{\left(\frac{3}{5}\right)^2 - x^2}}$. This fits the standard template $\int \frac{dx}{\sqrt{a^2 - x^2}}$ where $a = \frac{3}{5}$. Step 3: Apply the formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a} ight) + C$. Step 4: Substitute $a = \frac{3}{5}$ into the formula: $\frac{1}{5} \sin^{-1}\left(\frac{x}{3/5}\right) + C = \frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C$. Final answer: $\frac{1}{5} \sin^{-1}\left(\frac{5x}{3} ight) + C$
• Q: Find $\int \frac{x+2}{x^2 + 4x + 5} dx$. A: Step 1: Differentiate the denominator $x^2+4x+5$ to get $2x+4 = 2(x+2)$. Step 2: Express the numerator in terms of this derivative: $x+2 = \frac{1}{2}(2x+4)$. Step 3: Rewrite the integral as $\int \frac{\frac{1}{2}(2x+4)}{x^2+4x+5} dx = \frac{1}{2} \int \frac{2x+4}{x^2+4x+5} dx$. Step 4: Let $t = x^2+4x+5 \implies dt = (2x+4)dx$. The integral reduces to $\frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log|t| + C$. Step 5: Substitute $t = x^2+4x+5$ back into the expression. Final answer: $\frac{1}{2} \log|x^2+4x+5| + C$

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