Daily Integral 12 February 2026
The beginner and easy are both fine today, but damn what the hell is the medium integral. Got some help on discord and friends to do it.
I = ∫ 1 2 e 1 x x 2 d x Let u = 1 x , d u d x = − x − 2 I = ∫ 1 2 1 e u d u = [ e u ] 1 2 1 = e − e 1 2 \begin{gathered}
I = \int_1^2 \frac{e^{\frac{1}{x}}}{x^2}\ dx\\
\text{Let $u = \frac{1}{x}$, $\frac{du}{dx} = -x^{-2}$} \\
I = \int_\frac{1}{2}^1 e^u\ du \\
=\left[e^u\right]^1_\frac{1}{2} \\
=\boxed{e-e^\frac{1}{2}}
\end{gathered}
I = ∫ 1 2 x 2 e x 1 d x Let u = x 1 , d x d u = − x − 2 I = ∫ 2 1 1 e u d u = [ e u ] 2 1 1 = e − e 2 1
I = ∫ 0 1 l o g 2 ( 2 x + 4 x ) − l o g 2 ( 1 + 2 x ) 1 ( 1 + e − x ) 2 + 1 ( 1 + e x ) 2 + 2 e x + 2 + e − x Numerator = l o g 2 ( 2 x + 4 x ) − l o g 2 ( 1 + 2 x ) = l o g 2 ( 2 x + 4 x 1 + 2 x ) = l o g 2 ( 2 x ( 1 + 2 x ) 1 + 2 x ) = l o g 2 ( 2 x ) = x Denominator = 1 ( 1 + e − x ) 2 + 1 ( 1 + e x ) 2 + 2 e x + 2 + e − x = 1 1 + 2 e − x + e − 2 x + 1 1 + 2 e x + e 2 x + 2 e x + 2 + e − x = e x e x + 2 + e − x + e − x e − x + 2 + e x + 2 e x + 2 + e − x = e x + e − x + 2 e x + 2 + e − x = 1 I = ∫ 0 1 x d x [ 1 2 x 2 ] 0 1 = 1 2 \begin{gathered}
I = \int_0^1 \frac{log_2(2^x + 4^x) - log_2(1 + 2^x)}{\sqrt{\frac{1}{(1 + e^{-x})^2} + \frac{1}{(1 + e^x)^2}+\frac{2}{e^x + 2 + e^{-x}}}} \\
\ \\
\begin{aligned}
\text{Numerator}\ &= log_2(2^x + 4^x) - log_2(1 + 2^x) \\
&= log_2\left(\frac{2^x + 4^x}{1 + 2^x}\right) \\
&= log_2\left(\frac{2^x(1 + 2^x)}{1 + 2^x}\right) \\
&= log_2\left(2^x\right) \\
&=x
\end{aligned}
\ \\
\begin{aligned}
\text{Denominator} &= \sqrt{\frac{1}{(1 + e^{-x})^2} + \frac{1}{(1 + e^x)^2}+\frac{2}{e^x + 2 + e^{-x}}} \\
&= \sqrt{\frac{1}{1+2e^{-x}+e^{-2x}} + \frac{1}{1 + 2e^x + e^{2x}}+\frac{2}{e^x + 2 + e^{-x}}} \\
&= \sqrt{\frac{e^x}{e^x+2+e^{-x}} + \frac{e^{-x}}{e^{-x} + 2 + e^x}+\frac{2}{e^x + 2 + e^{-x}}} \\
&= \sqrt{\frac{e^x + e^{-x} + 2}{e^x + 2 + e^{-x}}} \\
&= 1
\end{aligned}
\ \\
I = \int_0^1 x\ dx \\
\left[\frac{1}{2}x^2\right]_0^1 = \boxed{\frac{1}{2}}
\end{gathered}
I = ∫ 0 1 ( 1 + e − x ) 2 1 + ( 1 + e x ) 2 1 + e x + 2 + e − x 2 l o g 2 ( 2 x + 4 x ) − l o g 2 ( 1 + 2 x ) Numerator = l o g 2 ( 2 x + 4 x ) − l o g 2 ( 1 + 2 x ) = l o g 2 ( 1 + 2 x 2 x + 4 x ) = l o g 2 ( 1 + 2 x 2 x ( 1 + 2 x ) ) = l o g 2 ( 2 x ) = x Denominator = ( 1 + e − x ) 2 1 + ( 1 + e x ) 2 1 + e x + 2 + e − x 2 = 1 + 2 e − x + e − 2 x 1 + 1 + 2 e x + e 2 x 1 + e x + 2 + e − x 2 = e x + 2 + e − x e x + e − x + 2 + e x e − x + e x + 2 + e − x 2 = e x + 2 + e − x e x + e − x + 2 = 1 I = ∫ 0 1 x d x [ 2 1 x 2 ] 0 1 = 2 1
I = ∫ − ∞ 0 e x e x + e x e x + e x e x + ⋯ d x Let y = e x + e x e x + e x ⋯ y = u + u y y 2 = u y + u u = y 2 y + 1 d u = 2 y ( y + 1 ) − y 2 ( y + 1 ) 2 d y u = 1 ⟹ y + 1 = y 2 y 2 − y − 1 = 0 y = 1 ± 5 2 Reject y = 1 − 5 2 , since e x > 0 , ∀ x ∈ R ∴ y = ϕ u = 0 ⟹ y = 0 Sub into original integrand I = ∫ 0 ϕ 1 y ⋅ 2 y ( y + 1 ) − y 2 ( y + 1 ) 2 d y = ∫ 0 ϕ 1 y ⋅ 2 y 2 + 2 y − y 2 ( y + 1 ) 2 d y = ∫ 0 ϕ y + 2 ( y + 1 ) 2 d y = ∫ 0 ϕ 1 y + 1 + 1 ( y + 1 ) 2 d y = [ l n ∣ y + 1 ∣ − 1 y + 1 ] 0 ϕ = l n ( ϕ + 1 ) − 1 ϕ + 1 + 1 \begin{gathered}
I = \int_{-\infty}^0\cfrac{e^x}{e^x + \cfrac{e^x}{e^x + \cfrac{e^x}{e^x + \cdots}}}\ dx \\
\ \\
\text{Let $y = e^x + \cfrac{e^x}{e^x + \cfrac{e^x}{\cdots}}$} \\
y = u + \frac{u}{y} \\
y^2 = uy + u \\
u = \frac{y^2}{y + 1} \\
du = \frac{2y(y + 1) - y^2}{(y+1)^2}\ dy\\
\ \\
u = 1 \implies y + 1 = y^2 \\
y^2 - y - 1 = 0 \\
y = \frac{1 ± \sqrt{5}}{2} \\
\text{Reject $y = \frac{1 - \sqrt{5}}{2}$, since $e^x > 0$, $\forall\ x\in \mathbb{R}$} \\
\therefore y = \phi \\
\ \\
u = 0 \implies y = 0
\ \\
\text{Sub into original integrand} \\
I = \int_0^\phi \frac{1}{y}\cdot\frac{2y(y+1)-y^2}{(y+1)^2}\ dy \\
= \int_0^\phi \frac{1}{y}\cdot\frac{2y^2+2y-y^2}{(y+1)^2}\ dy \\
= \int_0^\phi\frac{y+2}{(y+1)^2} dy \\
= \int_0^\phi \frac{1}{y + 1} + \frac{1}{(y + 1)^2} dy \\
= \left[ln|y + 1| - \frac{1}{y + 1}\right]_0^\phi \\
= \boxed{ln(\phi + 1) - \frac{1}{\phi + 1} + 1}
\end{gathered}
I = ∫ − ∞ 0 e x + e x + e x + ⋯ e x e x e x d x Let y = e x + e x + ⋯ e x e x y = u + y u y 2 = u y + u u = y + 1 y 2 d u = ( y + 1 ) 2 2 y ( y + 1 ) − y 2 d y u = 1 ⟹ y + 1 = y 2 y 2 − y − 1 = 0 y = 2 1 ± 5 Reject y = 2 1 − 5 , since e x > 0, ∀ x ∈ R ∴ y = ϕ u = 0 ⟹ y = 0 Sub into original integrand I = ∫ 0 ϕ y 1 ⋅ ( y + 1 ) 2 2 y ( y + 1 ) − y 2 d y = ∫ 0 ϕ y 1 ⋅ ( y + 1 ) 2 2 y 2 + 2 y − y 2 d y = ∫ 0 ϕ ( y + 1 ) 2 y + 2 d y = ∫ 0 ϕ y + 1 1 + ( y + 1 ) 2 1 d y = [ l n ∣ y + 1 ∣ − y + 1 1 ] 0 ϕ = l n ( ϕ + 1 ) − ϕ + 1 1 + 1
