1
1 1
1
x² – a²
· If
tal
]
then find the value of K.
>
Kx-a
x + a
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0
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Answered by
0
Answer
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
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