Question

evaluate integral 0 to 1
$( {x}^{2} + a)dx$
​

Answer 1

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