11)
Write:
\int e^x\left(lnx+\frac{1}{x^2}\right)dx=\int e^x\left(lnx+\frac{1}{x}\right)dx+\int e^x\left(-\frac{1}{x}+\frac{1}{x^2}\right)dx
Q1
Statement 1: \int e^x(log_{e}x + \frac{1}{x^2}) dx = e^x (log_{e}x - \frac{1}{x}) + C
Statement 2: \int e^x (f(x) + f^{ '}(x)) dx = e^x f(x)+ C
Answer: Both are true and 1 is correct explanation for 2
But shouldn't Statement 1 be wrong? Because f(x) in this case is log_{e}x and it's differentiation is 1/x not 1/x2
Please explain!
Q2: \int \frac{dx}{5+4sinx}
Hint: Take \int \frac{dx}{5+4sinx} = \frac{2tanx}{1+tan^2 x} .
I tried to solve it, but stucked later.
Q3: \int \frac{(x-1)dx}{(x+1)\sqrt{x^3 + x^2 + x}}
-
UP 0 DOWN 0 0 6
6 Answers
1
1Q2. u hav to take sinx = 2tan x1 + tan2x not as u mentioned in HINT
1Q3)
\int \frac{(x-1)}{(x+1)\sqrt{x^{3}+x^{2}+x}}dx
\int \frac{(x^{2}-1)}{(x+1)^{2}\sqrt{x^{3}+x^{2}+x}}dx
\int \frac{(x^{2}-1)}{(x^{2}+2x+1)\sqrt{x^{3}+x^{2}+x}}dx
\int \frac{(x^{2}-1)}{x.x(x+2+\frac{1}{x})\sqrt{x+1+\frac{1}{x}}}dx
\int \frac{(1-\frac{1}{x^{2}})}{(x+2+\frac{1}{x})\sqrt{x+1+\frac{1}{x}}}dx [ the one that was x2 in deno. divide it in the nume. ]
Put x + 1 + 1x = t
→ 1 - 1x2dx = dt
\Rightarrow \int \frac{dt}{(t+1)\sqrt{t}}
put t = u2 → dt = 2u du
\Rightarrow \int \frac{2udu}{(u^{2}+1)u}\Rightarrow 2\int \frac{du}{1+u^{2}}
= 2tan^{-1}u+c=2tan^{-1}\sqrt{x+1+\frac{1}{x}}+c
112) Write the integral as
\int \frac{(sin^2\frac{x}{2}+cos^2\frac{x}{2})dx}{5(sin^2\frac{x}{2}+cos^2\frac{x}{2})+8sin\frac{x}{2}cos\frac{x}{2}}
Dvide numerator and denom by cos^2\frac{x}{2} then take tan\frac{x}{2}=k.
That's enough - I suppose.
39Abey yaar... you posted this in my cb aur mujhe kuch karne ka mauka tak nahi diya!! LOL