Tolong bantu yaa kakk soal integral ini

<p>1. ∫axa−1dx\int a x^{a-1} dx</p><p>∫xndx=xn+1n+1+C(dengan&nbsp;n≠−1).\int x^n dx = \frac{x^{n+1}}{n+1} + C \quad \text{(dengan } n \neq -1\text{)}. ∫axa−1dx=a⋅xaa+C=xa+C.\int a x^{a-1} dx = a \cdot \frac{x^a}{a} + C = x^a + C.&nbsp;</p><p>2. ∫(2x3−1)dx\int (2x^3 - 1) dx</p><p>∫(2x3−1)dx=∫2x3dx−∫1dx.\int (2x^3 - 1) dx = \int 2x^3 dx - \int 1 dx. =2⋅x44−x+C=x42−x+C.= 2 \cdot \frac{x^4}{4} - x + C = \frac{x^4}{2} - x + C.&nbsp;</p><p>3. ∫(4x2−1)(4x3−3x)3dx\int (4x^2 - 1)(4x^3 - 3x)^3 dx</p><p>Ubah bentuk integral ini dengan substitusi:<br>Misalkan u=4x3−3xu = 4x^3 - 3x, maka turunan uu adalah:</p><p>dudx=12x2−3  ⟹  du=(4x2−1)dx.\frac{du}{dx} = 12x^2 - 3 \implies du = (4x^2 - 1) dx.</p><p>Integral menjadi:</p><p>∫(4x2−1)(4x3−3x)3dx=∫u3du=u44+C.\int (4x^2 - 1)(4x^3 - 3x)^3 dx = \int u^3 du = \frac{u^4}{4} + C.</p><p>Kembalikan uu:</p><p>(4x3−3x)44+C.\frac{(4x^3 - 3x)^4}{4} + C.&nbsp;</p><p>4. ∫x2(x2+1)3dx\int x^2 (x^2 + 1)^3 dx</p><p>Gunakan substitusi:<br>Misalkan u=x2+1u = x^2 + 1, maka turunan uu adalah:</p><p>dudx=2x  ⟹  xdx=12du.\frac{du}{dx} = 2x \implies x dx = \frac{1}{2} du.</p><p>Integral menjadi:</p><p>∫x2(x2+1)3dx=∫(u−1)u3⋅12du.\int x^2 (x^2 + 1)^3 dx = \int (u - 1) u^3 \cdot \frac{1}{2} du.</p><p>Perluas:</p><p>=12∫(u4−u3)du.= \frac{1}{2} \int (u^4 - u^3) du.</p><p>Integralkan:</p><p>=12(u55−u44)+C.= \frac{1}{2} \left( \frac{u^5}{5} - \frac{u^4}{4} \right) + C.</p><p>Kembalikan uu:</p><p>12((x2+1)55−(x2+1)44)+C.\frac{1}{2} \left( \frac{(x^2 + 1)^5}{5} - \frac{(x^2 + 1)^4}{4} \right) + C.&nbsp;</p><p>&nbsp;</p>