When solving integrals with rational functions, substitution is often useful, especially when the denominator involves a quadratic expression. Pay close attention to how the limits of integration change when performing the substitution, as this ensures the integral remains properly evaluated. For terms like \( \frac{1}{u^2} \), remember that integrating powers of \( u \) involves basic power rule applications, and logarithmic integrals like \( \frac{1}{u} \) lead to natural logarithms. Always simplify the expression step by step for better clarity in your solution.
We start with the integral:
\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx \]
Let \( u = a + bx^2 \), so:
\[ du = 2bx \, dx \quad \text{and} \quad x \, dx = \frac{du}{2b}. \]
The limits change as follows:
For \( x = 0 \Rightarrow u = a \) and for \( x = 1 \Rightarrow u = a + b \).
Substitute into the integral:
\[ \int_{a}^{a+b} \frac{2a - u}{u^2} \cdot \frac{1}{2b} \, du. \]
Simplify:
\[ \frac{1}{2b} \int_{a}^{a+b} \left( \frac{2a}{u^2} - \frac{1}{u} \right) du. \]
Solve each term: For \( \int_{a}^{a+b} \frac{2a}{u^2} du \):
\[ \int \frac{2a}{u^2} du = -\frac{2a}{u}. \]
Evaluate:
\[ -\frac{2a}{a + b} + \frac{2a}{a}. \]
For \( \int_{a}^{a+b} \frac{1}{u} du \):
\[ \ln(a + b) - \ln(a). \]
Combine results:
\[ \frac{1}{2b} \left( -\frac{2a}{a + b} + 2 - (\ln(a + b) - \ln(a)) \right). \]
Final simplification gives:
\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx = \frac{1}{a + b}. \]
We start with the integral:
\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx \]
Step 1: Use substitution \( u = a + bx^2 \):
\[ du = 2bx \, dx \quad \text{and} \quad x \, dx = \frac{du}{2b}. \]Step 2: Change the limits of integration:
For \( x = 0 \Rightarrow u = a \), and for \( x = 1 \Rightarrow u = a + b \).Step 3: Substitute into the integral:
\[ \int_{a}^{a+b} \frac{2a - u}{u^2} \cdot \frac{1}{2b} \, du. \]Step 4: Simplify the expression:
\[ \frac{1}{2b} \int_{a}^{a+b} \left( \frac{2a}{u^2} - \frac{1}{u} \right) du. \]Step 5: Solve each term:
For \( \int_{a}^{a+b} \frac{2a}{u^2} du \), we know: \[ \int \frac{2a}{u^2} du = -\frac{2a}{u}. \] Evaluating: \[ -\frac{2a}{a + b} + \frac{2a}{a}. \] For \( \int_{a}^{a+b} \frac{1}{u} du \), we get: \[ \ln(a + b) - \ln(a). \]Step 6: Combine results:
\[ \frac{1}{2b} \left( -\frac{2a}{a + b} + 2 - (\ln(a + b) - \ln(a)) \right). \]Step 7: Final simplification:
\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx = \frac{1}{a + b}. \]List I | List II | ||
A. | \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{7}{2}}x}{\sin^{\frac{7}{2}}+\cos^{\frac{7}{2}}}dx\) | I. | \(\frac{\pi}{4}-\frac{1}{2}\) |
B. | \(\int\limits_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx\) | II. | 0 |
C. | \(\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}x\cos x\ dx\) | III. | \(\frac{\pi}{4}\) |
D. | \(\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x\ dx\) | IV. | \(\frac{\pi^2}{4}\) |
List-I (Words) | List-II (Definitions) |
(A) Theocracy | (I) One who keeps drugs for sale and puts up prescriptions |
(B) Megalomania | (II) One who collects and studies objects or artistic works from the distant past |
(C) Apothecary | (III) A government by divine guidance or religious leaders |
(D) Antiquarian | (IV) A morbid delusion of one’s power, importance or godliness |