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Answers
EXPLANATION.
⇒ ∫2ˣ + 3ˣ/5ˣ dx.
As we know that,
⇒ ∫aˣdx = aˣ/㏒(a) + c = aˣ(㏒ₐe) + c.
Apply this formula in equation, we get.
⇒ ∫2ˣ + 3ˣ/5ˣ dx.
⇒ ∫2ˣ/5ˣ dx + ∫3ˣ/5ˣ dx.
We can also write as,
⇒ ∫(2/5)ˣ dx + ∫(3/5)ˣ dx.
⇒ (2/5)ˣ/㏒(2/5) + (3/5)ˣ/㏒(3/5) + c.
MORE INFORMATION.
Answer:
EXPLANATION.
⇒ ∫2ˣ + 3ˣ/5ˣ dx.
As we know that,
⇒ ∫aˣdx = aˣ/㏒(a) + c = aˣ(㏒ₐe) + c.
Apply this formula in equation, we get.
⇒ ∫2ˣ + 3ˣ/5ˣ dx.
⇒ ∫2ˣ/5ˣ dx + ∫3ˣ/5ˣ dx.
We can also write as,
⇒ ∫(2/5)ˣ dx + ∫(3/5)ˣ dx.
⇒ (2/5)ˣ/㏒(2/5) + (3/5)ˣ/㏒(3/5) + c.
MORE INFORMATION.
\sf \implies \int \sqrt{a^{2} - x^{2} } dx = \dfrac{x}{2} \sqrt{a^{2} - x^{2} } + \dfrac{a^{2} }{2} sinx^{-1} \dfrac{x}{a} + c⟹∫
a
2
−x
2
dx=
2
x
a
2
−x
2
+
2
a
2
sinx
−1
a
x
+c
\sf \implies \int \sqrt{x^{2} + a^{2} } dx = \dfrac{x}{2} \sqrt{x^{2} + a^{2} } + \dfrac{a^{2} }{2}. sinh^{-1} \dfrac{x}{a} + c⟹∫
x
2
+a
2
dx=
2
x
x
2
+a
2
+
2
a
2
.sinh
−1
a
x
+c
\sf \implies \int \sqrt{x^{2} - a^{2} } dx = \dfrac{x}{2} \sqrt{x^{2} - a^{2} } - \dfrac{a^{2} }{2} .cosh^{-1} \dfrac{x}{a} + c.⟹∫
x
2
−a
2
dx=
2
x
x
2
−a
2
−
2
a
2
.cosh
−1
a
x
+c.
\sf \implies \int \dfrac{1}{x\sqrt{x^{2} - a^{2} } } dx = \dfrac{1}{a} sec^{-1} \dfrac{x}{a} + c.⟹∫
x
x
2
−a
2
1
dx=
a
1
sec
−1
a
x
+c.
\sf \implies \int e^{ax} sin bx dx = \dfrac{e^{ax} }{a^{2}+ b^{2} } (a sinbx - b cosbx) + c = \dfrac{e^{ax} }{\sqrt{a^{2} + b^{2} } } sin \bigg(bx - tan^{-1}\dfrac{b}{a}\bigg) + c⟹∫e
ax
sinbxdx=
a
2
+b
2
e
ax
(asinbx−bcosbx)+c=
a
2
+b
2
e
ax
sin(bx−tan
−1
a
b
)+c
\sf \implies \int e^{ax} cos bx dx = \dfrac{e^{ax} }{a^{2}+ b^{2} } (a cos bx + b sin bx ) + c = \dfrac{e^{ax} }{\sqrt{a^{2} + b^{2} } } cos \bigg(bx - tan^{-1} \dfrac{b}{a} \bigg) + c⟹∫e
ax
cosbxdx=
a
2
+b
2
e
ax
(acosbx+bsinbx)+c=
a
2
+b
2
e
ax
cos(bx−tan
−1
a
b
)+c