Question

Find the integral of 1/x^2+a^2 w.r.t x and hence find the 1/x^2_16 dx​

Answer 1

Answer:

Hope it is helpfull

Step-by-step explanation:

Given I=∫  

x  

2

−a  

2

 

​  

 

1

​  

dx

x=asecθ,dx=asecθtanθdθ

x  

2

−a  

2

 

​  

=atanθ

I=∫  

atanθ

asecθtanθdθ

​  

 

I=∫tanθdθ=ln∣secθ∣+c

I=∫  

atanθ

asecθtanθdθ

​  

 

I=∫secθdθ=ln∣secθ+tanθ∣+c

I=ln(  

a

x

​  

+  

(  

a

x

​  

)  

2

−1

​  

)+c  

I  

1

​  

=∫  

x  

2

+6x−7

​  

 

dx

​  

=∫  

(x+3)  

2

−4  

2

 

​  

 

dx

​  

 

x=x+3,a=4

I=ln  

⎝

⎛

​  

 

4

x+3

​  

+  

(  

4

x+3

​  

)  

2

−1

​  

 

⎠

⎞

​  

+c