⭐ Contoh Soal 1

Hitunglah:∫4x+1 dx\int \sqrt{4x+1} \, dx∫4x+1​dx

Penyelesaian (Substitusi)

Misal:u=4x+1u = 4x+1u=4x+1 du=4 dx⇒dx=14dudu = 4\, dx \quad \Rightarrow \quad dx = \frac{1}{4} dudu=4dx⇒dx=41​du

Maka:∫4x+1 dx=∫u⋅14 du\int \sqrt{4x+1}\, dx = \int \sqrt{u} \cdot \frac{1}{4}\, du∫4x+1​dx=∫u​⋅41​du =14∫u1/2 du= \frac{1}{4} \int u^{1/2}\, du=41​∫u1/2du =14⋅23u3/2+C= \frac{1}{4} \cdot \frac{2}{3}u^{3/2} + C=41​⋅32​u3/2+C =16(4x+1)3/2+C= \frac{1}{6}(4x+1)^{3/2} + C=61​(4x+1)3/2+C

✅ Jawaban:16(4x+1)3/2+C\boxed{\frac{1}{6}(4x+1)^{3/2} + C}61​(4x+1)3/2+C​

Baca juga:Contoh Soal Fisika Parabola: Rumus, Konsep, dan Pembahasan Lengk

⭐ Contoh Soal 2

Hitunglah:∫xx2+9 dx\int \frac{x}{\sqrt{x^2+9}}\, dx∫x2+9​x​dx

Penyelesaian

Substitusi:u=x2+9u = x^2 + 9u=x2+9 du=2x dx⇒x dx=12dudu = 2x\, dx \Rightarrow x\, dx = \frac{1}{2} dudu=2xdx⇒xdx=21​du

Maka:∫xx2+9 dx=∫1u⋅12du\int \frac{x}{\sqrt{x^2+9}}\, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2}du∫x2+9​x​dx=∫u​1​⋅21​du =12∫u−1/2 du= \frac{1}{2}\int u^{-1/2}\, du=21​∫u−1/2du =12⋅2u1/2+C= \frac{1}{2} \cdot 2u^{1/2} + C=21​⋅2u1/2+C =x2+9+C= \sqrt{x^2 + 9} + C=x2+9​+C

✅ Jawaban:x2+9+C\boxed{\sqrt{x^2+9} + C}x2+9​+C​

⭐ Contoh Soal 3

Hitunglah:∫x5x2−2 dx\int x\sqrt{5x^2 – 2}\, dx∫x5x2−2​dx

Penyelesaian

Substitusi:u=5x2−2u = 5x^2 – 2u=5×2−2 du=10x dx⇒x dx=110dudu = 10x\, dx \quad \Rightarrow \quad x\, dx = \frac{1}{10}dudu=10xdx⇒xdx=101​du

Maka:∫x5x2−2 dx=∫u⋅110 du\int x\sqrt{5x^2 – 2}\, dx = \int \sqrt{u} \cdot \frac{1}{10}\, du∫x5x2−2​dx=∫u​⋅101​du =110∫u1/2 du= \frac{1}{10}\int u^{1/2}\, du=101​∫u1/2du =110⋅23u3/2+C= \frac{1}{10} \cdot \frac{2}{3}u^{3/2} + C=101​⋅32​u3/2+C =115(5x2−2)3/2+C= \frac{1}{15}(5x^2 – 2)^{3/2} + C=151​(5×2−2)3/2+C

✅ Jawaban:115(5x2−2)3/2+C\boxed{\frac{1}{15}(5x^2 – 2)^{3/2} + C}151​(5×2−2)3/2+C​

Baca juga :FEB Universitas Teknokrat Indonesia Gelar Kuliah Umum Ekonomi Mikro Bahas Peran Pemerintah dalam Penyediaan Barang Publik

⭐ Contoh Soal 4

Hitunglah:∫dx3−2x\int \frac{dx}{\sqrt{3 – 2x}}∫3−2x​dx​

Penyelesaian

Substitusi:u=3−2xu = 3 – 2xu=3−2x du=−2dx⇒dx=−12dudu = -2 dx \Rightarrow dx = -\frac{1}{2} dudu=−2dx⇒dx=−21​du ∫dx3−2x=−12∫u−1/2 du\int \frac{dx}{\sqrt{3 – 2x}} = -\frac{1}{2}\int u^{-1/2}\, du∫3−2x​dx​=−21​∫u−1/2du =−12⋅2u1/2+C= -\frac{1}{2} \cdot 2u^{1/2} + C=−21​⋅2u1/2+C =−3−2x+C= -\sqrt{3 – 2x} + C=−3−2x​+C

Penulis:loveytha

Leave a Reply

Your email address will not be published. Required fields are marked *