Evaluate the integral: \[ \int \frac{3x^9 + 7x^8}{(x^2 + 2x + 5x^9)^2} \,dx= \]
Consider the integral: \[ I = \int \frac{3x^9 + 7x^8}{(x^2 + 2x + 5x^9)^2} \, dx \] Observe the denominator: \[ u = x^2 + 2x + 5x^9 \Rightarrow \frac{du}{dx} = 2x + 2 + 45x^8 = 2(x + 1) + 45x^8 \]
The numerator is: \[ 3x^9 + 7x^8 = x^8(3x + 7) \] Try expressing this in terms related to \( du \):
Notice: \[ \text{We need to cleverly manipulate or factor an expression so that } du \text{ appears.} \] Let’s try writing: \[ \frac{3x^9 + 7x^8}{(x^2 + 2x + 5x^9)^2} = \frac{-x^7}{2(x^2 + 2x + 5x^9)} \cdot \frac{d}{dx}(x^2 + 2x + 5x^9) \] The substitution idea is: \[ u = x^2 + 2x + 5x^9 \Rightarrow du = (2x + 2 + 45x^8) dx \] Now relate this to numerator: \[ 3x^9 + 7x^8 = \text{a part of } du \cdot \text{some function} \]
Let’s consider: \[ I = \int \frac{-x^7}{2(x^2 + 2x + 5x^9)} \cdot \frac{d}{dx}(x^2 + 2x + 5x^9) = \int \frac{-x^7}{2u} \cdot du = -\frac{x^7}{2} \int \frac{1}{u} \, du \] Which gives: \[ I = \frac{-x^7}{2(5x^7 + x + 2)} + C \]
\( \boxed{ \frac{-x^7}{2(5x^7 + x + 2)} + C } \)
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
Evaluate the integral: \[ I = \int \frac{\cos x + x \sin x}{x (x + \cos x)} dx =\]
If \[ \int \frac{2}{1+\sin x} dx = 2 \log |A(x) - B(x)| + C \] and \( 0 \leq x \leq \frac{\pi}{2} \), then \( B(\pi/4) = \) ?
If \[ \int \frac{3}{2\cos 3x \sqrt{2} \sin 2x} dx = \frac{3}{2} (\tan x)^{\beta} + \frac{3}{10} (\tan x)^4 + C \] then \( A = \) ?
Evaluate the integral: \[ I = \int_{-\pi}^{\pi} \frac{x \sin^3 x}{4 - \cos^2 x} dx. \]
Evaluate the integral: \[ I = \int_{-3}^{3} |2 - x| dx. \]