pls answer correctly with reasons.
Answers
Answer:
1. Given to evaluate,
\displaystyle\longrightarrow I=\int\dfrac{dx}{x^2-16}⟶I=∫
x
2
−16
dx
\displaystyle\longrightarrow I=\int\dfrac{dx}{(x-4)(x+4)}⟶I=∫
(x−4)(x+4)
dx
Multiplying and dividing the integrand by 4 - (- 4) = 8,
\displaystyle\longrightarrow I=\dfrac{1}{8}\int\dfrac{8}{(x-4)(x+4)}\ dx⟶I=
8
1
∫
(x−4)(x+4)
8
dx
\displaystyle\longrightarrow I=\dfrac{1}{8}\int\dfrac{(x+4)-(x-4)}{(x-4)(x+4)}\ dx⟶I=
8
1
∫
(x−4)(x+4)
(x+4)−(x−4)
dx
\displaystyle\longrightarrow I=\dfrac{1}{8}\int\left(\dfrac{1}{x-4}-\dfrac{1}{x+4}\right)\ dx⟶I=
8
1
∫(
x−4
1
−
x+4
1
) dx
\displaystyle\longrightarrow I=\dfrac{1}{8}\big[\log|x-4|-\log|x+4|\big]+C⟶I=
8
1
[log∣x−4∣−log∣x+4∣]+C
\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{8}\log\left|\dfrac{x-4}{x+4}\right|+C}}⟶
I=
8
1
log
∣
∣
∣
∣
∣
x+4
x−4
∣
∣
∣
∣
∣
+C
2. Given to evaluate,
\displaystyle\longrightarrow I=\int\dfrac{dx}{2x-x^2}⟶I=∫
2x−x
2
dx
\displaystyle\longrightarrow I=\int\dfrac{dx}{x(2-x)}⟶I=∫
x(2−x)
dx
or,
\displaystyle\longrightarrow I=-\int\dfrac{dx}{(x+0)(x-2)}⟶I=−∫
(x+0)(x−2)
dx
Multiplying and dividing the integrand by 0 - (- 2) = 2,
\displaystyle\longrightarrow I=-\dfrac{1}{2}\int\dfrac{2}{x(x-2)}\ dx⟶I=−
2
1
∫
x(x−2)
2
dx
\displaystyle\longrightarrow I=-\dfrac{1}{2}\int\dfrac{x-(x-2)}{x(x-2)}\ dx⟶I=−
2
1
∫
x(x−2)
x−(x−2)
dx
\displaystyle\longrightarrow I=-\dfrac{1}{2}\int\left(\dfrac{1}{x-2}-\dfrac{1}{x}\right)\ dx⟶I=−
2
1
∫(
x−2
1
−
x
1
) dx
\displaystyle\longrightarrow I=-\dfrac{1}{2}\big[\log|x-2|-\log|x|\big]+C⟶I=−
2
1
[log∣x−2∣−log∣x∣]+C
\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{2}\log\left|\dfrac{x}{x-2}\right|+C}}⟶
I=
2
1
log
∣
∣
∣
∣
∣
x−2
x
∣
∣
∣
∣
∣
+C