Sertakan cara terbaiknya ya... :).... Question from @B3ST

ridhovictor

Jawab:

1. Modifikasi x

2. Subtitusi 1

3. Parsial

4. Sub trig

5.Subtitusi 2

6.Pitagoras

7.Selesai

Penjelasan dengan langkah-langkah:

2 gambar terakhir di jawaban selanjutnya

7 votes Thanks 9

ridhovictor https://drive.google.com/file/d/1-qRzrvLku6lH03w98CuqsJJIiT4IsEaC/view?usp=sharing

dinda7014 Assalamualaikum ka

Adjie564 Kalau menurut saya sih gini :

$\int{\left(\frac{\sqrt{x^4+2x^3}}{x^4}\right)\,dx}=\int{\left(\frac{\sqrt{x^3}.\sqrt{x+2}}{x^4}\right)\,dx}\\\int{\frac{\sqrt{x+2}}{x^2\sqrt{x}}\,dx}$

Substitusi :

u = x + 2 ----> x = u - 2

du = dx

maka :

$\int{\frac{x+2}{x^2\sqrt{x}}\,dx}=\int{\frac{\sqrt{u}}{{(u-2)}^{2}\sqrt{u-2}}\,du}\cdots(1)$

Substitusi 2 :

$t=\sqrt{u-2}\\t^2=u-2\\u=t^2+2\\du=2t\,dt$

maka :

$\int{\frac{\sqrt{u}}{{(u-2)}^{2}\sqrt{u-2}}\,du}=\int{\frac{\sqrt{t^2+2}}{t^4\sqrt{t^2}}\,(2t)\,dt}\\\int{\frac{2\sqrt{t^2+2}}{t^4}\,dt}\cdots(2)$

Substitusi Trigonometri :

$t=\sqrt{2}\tan{a}\to\,\tan{a}=\frac{t}{2}\sqrt{2}\\dt=\sqrt{2}\sec^2{a}\,da$

maka :

$\int{\frac{2\sqrt{t^2+2}}{t^4}\,dt}=\int{\frac{2\sqrt{2\tan^2{a}+2}}{4\tan^4{a}}\,(\sec^2{a}\,da)}\\\int{\frac{2\sqrt{2}\sec^3{a}}{4\tan^4{a}}\,da=\frac{1}{2}\sqrt{2}\int{\frac{\frac{1}{cos^3{a}}}{\frac{sin^4{a}}{cos^4{a}}}}\,da}=\frac{1}{2}\sqrt{2}\int{\frac{\cos{a}}{\sin^4{a}}\,da}\cdots(3)$

Substitusi 3 :

$b=\sin{a}\\db=\cos{a}\,da$

maka :

$\frac{1}{2}\sqrt{2}\int{\frac{\cos{a}}{\sin^4{a}}\,da}=\frac{1}{2}\sqrt{2}\int{\frac{db}{b^4}}=-\frac{1}{6}\sqrt{2}b^{-3}+c=-\frac{1}{6}\sqrt{2}\sin^{-3}{a}+c=-\frac{1}{6}\sqrt{2}\cosec^3{a}+c$

Dengan segitiga siku - siku, akan diperoleh :

$-\frac{1}{6}\sqrt{2}{\left(\frac{\sqrt{t^2+2}}{t}\right)}^{3}+c=-\frac{1}{6}\sqrt{2}{\left(\frac{\sqrt{x+2}}{\sqrt{x}}\right)}^{3}+c$

Maaf jika salah. Inilah cara terbaik saya.

4 votes Thanks 8

Tafsir dr ayat berikut :

Jika a² + a - 3 = 0 Berapa nilai dari a²(a + 4) + 2013 ?

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