Penjelasan dengan langkah-langkah:
Nomor 3 :
[tex]\tt \int\limits^1_0{\frac{x+2}{(x^2+4x+1)}} \, dx \\\\=\int\limits^1_0 {\frac{x+2}{(x^2+4x+1)^2}\times \frac{1}{2x+4}} \, dx \\\\=\int\limits^1_0 {\frac{1}{(x^2+4x+1)^2}\times \frac{1}{2}} \, dt\\\\= \int\limits^1_0 {\frac{1}{2(x^2+4x+1)^2}} \, dt\to gunakan~sifat~integral:\int\limits {\frac{1}{x^n}} \, dx=-\frac{1}{(n-1)x^{n-1}} \\\\= \frac{1}{2}(-\frac{1}{(2-1)(x^2+4x+1)^{2-1}})|^1_0\\ \\=\frac{1}{2}(-\frac{1}{x^2+4x+1})|^1_0\\ \\=-\frac{1}{2(1)^2+8(1)+2}-(-\frac{1}{2(0)^2+8(0)+2})[/tex]
[tex]\tt =\frac{5}{12}[/tex]
Nomor 4 :
[tex]\tt \int\limits {x(x+1)^2} \, dx \to gunakan~bentuk~pemfaktoran:(a+b)^2=a^2+2ab+b^2\\\\= \int\limits {x(x^2+2x+1)} \, dx \\\\=\frac{x^{3+1}}{3+1}+\frac{2x^{2+1}}{2+1}+\frac{x^{1+1}}{1+1}\\ \\ =\frac{x^4}{4}+\frac{2x^3}{3}+\frac{x^2}{2}+C[/tex]
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Penjelasan dengan langkah-langkah:
Nomor 3 :
[tex]\tt \int\limits^1_0{\frac{x+2}{(x^2+4x+1)}} \, dx \\\\=\int\limits^1_0 {\frac{x+2}{(x^2+4x+1)^2}\times \frac{1}{2x+4}} \, dx \\\\=\int\limits^1_0 {\frac{1}{(x^2+4x+1)^2}\times \frac{1}{2}} \, dt\\\\= \int\limits^1_0 {\frac{1}{2(x^2+4x+1)^2}} \, dt\to gunakan~sifat~integral:\int\limits {\frac{1}{x^n}} \, dx=-\frac{1}{(n-1)x^{n-1}} \\\\= \frac{1}{2}(-\frac{1}{(2-1)(x^2+4x+1)^{2-1}})|^1_0\\ \\=\frac{1}{2}(-\frac{1}{x^2+4x+1})|^1_0\\ \\=-\frac{1}{2(1)^2+8(1)+2}-(-\frac{1}{2(0)^2+8(0)+2})[/tex]
[tex]\tt =\frac{5}{12}[/tex]
Nomor 4 :
[tex]\tt \int\limits {x(x+1)^2} \, dx \to gunakan~bentuk~pemfaktoran:(a+b)^2=a^2+2ab+b^2\\\\= \int\limits {x(x^2+2x+1)} \, dx \\\\=\frac{x^{3+1}}{3+1}+\frac{2x^{2+1}}{2+1}+\frac{x^{1+1}}{1+1}\\ \\ =\frac{x^4}{4}+\frac{2x^3}{3}+\frac{x^2}{2}+C[/tex]