Apka solution dekhne ke bad aisa lg rha h ki wakay me integration itna easy method me solve kr skte h 😍😄 Thank u so much ma'am 🥰😊 kaas apke vdo 1 saal phle mil jate 😒🙏🙏❣️❣️
Steps divide by cos⁴x in num. / Denominator and then breck upper sec⁴x=sec²x.sec²x Than write one sec²x =(1+tan²x) then put t= tan x and then use partial fraction
@@TRIGU_LOVER Babu mai graduation 1st year me hu bsc maths honours from bhu integration hard nahi hai mehnat Karo sabse jada maja usi ko lagane me ayega aur daily kam se kam 5 se 10 questions lagao daily revise karte raho asani se ho jayega chahe kaisa bhi questions ho bas concept na bhulana
To evaluate the integral \( I = \int \frac{x^2 + 2}{x^2 + 1} \, dx \), we can use the method of splitting the fraction and simplifying. Here's the detailed process: 1. **Split the fraction**: \[ \frac{x^2 + 2}{x^2 + 1} = \frac{x^2 + 1 + 1}{x^2 + 1} = \frac{x^2 + 1}{x^2 + 1} + \frac{1}{x^2 + 1} \] This simplifies to: \[ \frac{x^2 + 2}{x^2 + 1} = 1 + \frac{1}{x^2 + 1} \] 2. **Separate the integral**: \[ I = \int \left(1 + \frac{1}{x^2 + 1} ight) \, dx \] This can be written as two separate integrals: \[ I = \int 1 \, dx + \int \frac{1}{x^2 + 1} \, dx \] 3. **Integrate each part**: - The integral of 1 with respect to \( x \) is \( x \): \[ \int 1 \, dx = x \] - The integral of \( \frac{1}{x^2 + 1} \) with respect to \( x \) is \( \arctan(x) \): \[ \int \frac{1}{x^2 + 1} \, dx = \arctan(x) \] 4. **Combine the results**: \[ I = x + \arctan(x) + C \] where \( C \) is the constant of integration. So, the final result is: \[ I = x + \arctan(x) + C \] Is this correct
Why are you using latex here Btw your answer is correct Just write 1+1 in place of 2 Then write the integrals separately i, e integral x²+1+1/x²+2 = integral x²+1/x²+1 + integral 1/x²+1 =integral 1dx + tan inverse x/1(because standard integral) =x + tan inverse x You can solve this in ur mind without using a pen. Thank you
