⭐ Contoh Soal 1

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

Penyelesaian (Substitusi)

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

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

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

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

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⭐ Contoh Soal 2

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

Penyelesaian

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

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

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

⭐ Contoh Soal 3

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

Penyelesaian

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

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

✅ Jawaban:$$\boxed{\frac{1}{15}(5x^2 – 2)^{3/2} + C}$$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:$$\int \frac{dx}{\sqrt{3 – 2x}}$$∫3−2x​dx​

Penyelesaian

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

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