Jawab:
Saya jawab dari yang termudah
Penjelasan dengan langkah-langkah:
Nomor 1 [tex]\displaystyle \int (ax+b)^n~dx[/tex]
Integral substitusi
[tex]\displaystyle u=ax+b\\du=a~dx[/tex]
[tex]\begin{aligned}\int (ax+b)^n&\:=\int u^n~\frac{du}{a}\\\:&=\frac{u^{n+1}}{a(n+1)}+C\\\:&=\frac{(ax+b)^{n+1}}{a(n+1)}+C\end{aligned}[/tex]
Nomor 2 [tex]\displaystyle \int \frac{dx}{(ax+b)^n}[/tex]
[tex]\begin{aligned}\int \frac{dx}{(ax+b)^n}&\:=\int \frac{1}{u^n}~\frac{du}{a}\\\:&=\int \frac{u^{-n}}{a}~du\\\:&=\int \frac{u^{-n+1}}{a(-n+1)}+C\\\:&=-\frac{1}{a(n-1)u^{n-1}}+C\\\:&=-\frac{1}{a(n-1)(ax+b)^{n-1}}+C\end{aligned}[/tex]
Nomor 3 [tex]\displaystyle \int e^{3x}~dx[/tex]
[tex]\begin{aligned}u&\:=3x\\du\:&=3~dx\end{aligned}[/tex]
[tex]\begin{aligned}\int e^{3x}~dx&\:=\int e^u~\frac{du}{3}\\\:&=\frac{1}{3}~e^u+C\\\:&=\frac{1}{3}~e^{3x}+C\end{aligned}[/tex]
Nomor 4 [tex]\displaystyle \int \frac{\cos x}{\sqrt{1+\sin x}}~dx[/tex]
[tex]\begin{aligned}u&\:=1+\sin x\\du\:&=\cos x~dx\end{aligned}[/tex]
[tex]\begin{aligned}\int \frac{\cos x}{\sqrt{1+\sin x}}&\:=\int \frac{\cancel{\cos x}}{\sqrt{u}}~\frac{du}{\cancel{\cos x}}\\\:&=\int u^{-\frac{1}{2}}~du\\\:&=\int 2u^{\frac{1}{2}}+C\\\:&=2\sqrt{u}+C\\\:&=2\sqrt{1+\sin x}+C\end{aligned}[/tex]
Nomor 5 [tex]\displaystyle \int \arctan 2x~dx[/tex]
Integral substitusi dan parsial
[tex]\displaystyle \int \arctan 2x~dx=\frac{1}{2}\int \arctan p~dp[/tex]
Selesaikan [tex]\displaystyle \int \arctan p~dp[/tex]
[tex]\displaystyle \begin{matrix}u=\arctan p & dv=dp\\ du=\frac{1}{1+p^2}~dp & v=p\end{matrix}[/tex]
[tex]\begin{aligned}\int u~dv&\:=uv-\int v~du\\\int \arctan p~dp\:&=p\arctan p-\int \frac{p}{1+p^2}~dp\\\:&=p\arctan p-\int \frac{p}{q}~\frac{dq}{2p}\\\:&=p\arctan p-\frac{1}{2}\ln|q|\\\:&=p\arctan p-\frac{\ln(1+p^2)}{2}\end{aligned}[/tex]
maka
[tex]\begin{aligned}\int \arctan 2x&\:=\frac{1}{2}\left [ p\arctan p-\frac{\ln(1+p^2)}{2} \right ]+C\\\:&=\frac{1}{2}\left [ 2x\arctan 2x-\frac{\ln(1+4x^2)}{2} \right ]+C\\\:&=x\arctan 2x-\frac{1}{4}\ln(4x^2+1)+C\end{aligned}[/tex]
Nomor 6 [tex]\displaystyle \int \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2}~dx[/tex]
Integral parsial
[tex]\displaystyle \int \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2}~dx=\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx-\int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx[/tex]
Selesaikan [tex]\displaystyle \int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx[/tex]
[tex]\displaystyle \begin{matrix}u=e^{\arctan x} & dv=\frac{2x}{(1+x^2)^2}~dx\\du=\frac{e^{\arctan x}}{1+x^2}~dx & v=-\frac{1}{1+x^2}\end{matrix}\\\int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx=-\frac{e^{\arctan x}}{1+x^2}+\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx[/tex]
Diperoleh
[tex]\begin{aligned}\int \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2}~dx&\:=\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx-\int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx\\\:&=\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx-\left [ -\frac{e^{\arctan x}}{1+x^2}+\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx \right ]\\\:&=\frac{e^{\arctan x}}{1+x^2}+C\end{aligned}[/tex]
Nomor 7 [tex]\displaystyle \int \left ( \sqrt{\tan x}+\sqrt{\cot x} \right )dx[/tex]
Melibatkan trik manipulasi
[tex]\displaystyle \int \left ( \sqrt{\tan x}+\sqrt{\cot x} \right )dx\\=\int \left ( \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} \right )dx\\=\int \left ( \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right )dx\\=\frac{\sqrt{2}}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}~dx\\=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{2\sin x\cos x+1-1}}~dx\\=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{2\sin x\cos x+1-(\sin^2 x+\cos^2 x)}}~dx[/tex]
[tex]\displaystyle =\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}~dx\\=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{1-u^2}}~\frac{du}{\cos x+\sin x}\\=\sqrt{2}\arcsin u+C\\=\sqrt{2}\sin^{-1}(\sin x-\cos x)+C[/tex]
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Jawab:
Saya jawab dari yang termudah
Penjelasan dengan langkah-langkah:
Nomor 1 [tex]\displaystyle \int (ax+b)^n~dx[/tex]
Integral substitusi
[tex]\displaystyle u=ax+b\\du=a~dx[/tex]
[tex]\begin{aligned}\int (ax+b)^n&\:=\int u^n~\frac{du}{a}\\\:&=\frac{u^{n+1}}{a(n+1)}+C\\\:&=\frac{(ax+b)^{n+1}}{a(n+1)}+C\end{aligned}[/tex]
Nomor 2 [tex]\displaystyle \int \frac{dx}{(ax+b)^n}[/tex]
Integral substitusi
[tex]\begin{aligned}\int \frac{dx}{(ax+b)^n}&\:=\int \frac{1}{u^n}~\frac{du}{a}\\\:&=\int \frac{u^{-n}}{a}~du\\\:&=\int \frac{u^{-n+1}}{a(-n+1)}+C\\\:&=-\frac{1}{a(n-1)u^{n-1}}+C\\\:&=-\frac{1}{a(n-1)(ax+b)^{n-1}}+C\end{aligned}[/tex]
Nomor 3 [tex]\displaystyle \int e^{3x}~dx[/tex]
Integral substitusi
[tex]\begin{aligned}u&\:=3x\\du\:&=3~dx\end{aligned}[/tex]
[tex]\begin{aligned}\int e^{3x}~dx&\:=\int e^u~\frac{du}{3}\\\:&=\frac{1}{3}~e^u+C\\\:&=\frac{1}{3}~e^{3x}+C\end{aligned}[/tex]
Nomor 4 [tex]\displaystyle \int \frac{\cos x}{\sqrt{1+\sin x}}~dx[/tex]
Integral substitusi
[tex]\begin{aligned}u&\:=1+\sin x\\du\:&=\cos x~dx\end{aligned}[/tex]
[tex]\begin{aligned}\int \frac{\cos x}{\sqrt{1+\sin x}}&\:=\int \frac{\cancel{\cos x}}{\sqrt{u}}~\frac{du}{\cancel{\cos x}}\\\:&=\int u^{-\frac{1}{2}}~du\\\:&=\int 2u^{\frac{1}{2}}+C\\\:&=2\sqrt{u}+C\\\:&=2\sqrt{1+\sin x}+C\end{aligned}[/tex]
Nomor 5 [tex]\displaystyle \int \arctan 2x~dx[/tex]
Integral substitusi dan parsial
[tex]\displaystyle \int \arctan 2x~dx=\frac{1}{2}\int \arctan p~dp[/tex]
Selesaikan [tex]\displaystyle \int \arctan p~dp[/tex]
[tex]\displaystyle \begin{matrix}u=\arctan p & dv=dp\\ du=\frac{1}{1+p^2}~dp & v=p\end{matrix}[/tex]
[tex]\begin{aligned}\int u~dv&\:=uv-\int v~du\\\int \arctan p~dp\:&=p\arctan p-\int \frac{p}{1+p^2}~dp\\\:&=p\arctan p-\int \frac{p}{q}~\frac{dq}{2p}\\\:&=p\arctan p-\frac{1}{2}\ln|q|\\\:&=p\arctan p-\frac{\ln(1+p^2)}{2}\end{aligned}[/tex]
maka
[tex]\begin{aligned}\int \arctan 2x&\:=\frac{1}{2}\left [ p\arctan p-\frac{\ln(1+p^2)}{2} \right ]+C\\\:&=\frac{1}{2}\left [ 2x\arctan 2x-\frac{\ln(1+4x^2)}{2} \right ]+C\\\:&=x\arctan 2x-\frac{1}{4}\ln(4x^2+1)+C\end{aligned}[/tex]
Nomor 6 [tex]\displaystyle \int \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2}~dx[/tex]
Integral parsial
[tex]\displaystyle \int \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2}~dx=\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx-\int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx[/tex]
Selesaikan [tex]\displaystyle \int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx[/tex]
[tex]\displaystyle \begin{matrix}u=e^{\arctan x} & dv=\frac{2x}{(1+x^2)^2}~dx\\du=\frac{e^{\arctan x}}{1+x^2}~dx & v=-\frac{1}{1+x^2}\end{matrix}\\\int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx=-\frac{e^{\arctan x}}{1+x^2}+\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx[/tex]
Diperoleh
[tex]\begin{aligned}\int \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2}~dx&\:=\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx-\int \frac{2xe^{\arctan x}}{(1+x^2)^2}~dx\\\:&=\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx-\left [ -\frac{e^{\arctan x}}{1+x^2}+\int \frac{e^{\arctan x}}{(1+x^2)^2}~dx \right ]\\\:&=\frac{e^{\arctan x}}{1+x^2}+C\end{aligned}[/tex]
Nomor 7 [tex]\displaystyle \int \left ( \sqrt{\tan x}+\sqrt{\cot x} \right )dx[/tex]
Melibatkan trik manipulasi
[tex]\displaystyle \int \left ( \sqrt{\tan x}+\sqrt{\cot x} \right )dx\\=\int \left ( \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} \right )dx\\=\int \left ( \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right )dx\\=\frac{\sqrt{2}}{\sqrt{2}}\int \frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}~dx\\=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{2\sin x\cos x+1-1}}~dx\\=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{2\sin x\cos x+1-(\sin^2 x+\cos^2 x)}}~dx[/tex]
[tex]\displaystyle =\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}~dx\\=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{1-u^2}}~\frac{du}{\cos x+\sin x}\\=\sqrt{2}\arcsin u+C\\=\sqrt{2}\sin^{-1}(\sin x-\cos x)+C[/tex]