Home/Calculus II/A.10. Integrals with a+bua + bua+bu/FormulaA.10. Integrals with \(a + bu\)FormulaIn A.10. Integrals with a+bua + bua+buFormulaCopy LaTeX∫duu(a+bu)2=1a(a+bu)−1a2 ln∣a+buu∣+C\int \frac{du}{u(a + bu)^2} = \frac{1}{a(a + bu)} - \frac{1}{a^2}\,\ln\left|\frac{a + bu}{u}\right| + C∫u(a+bu)2du=a(a+bu)1−a21lnua+bu+CTagsA 10 integrals with A buDifferential calculusDerivativeDerivative formulaTableVariablesCCCConstante real de integraciónuuuVariable auxiliar de sustituciónaaaConstante real (a menudo a>0a>0a>0)dududuDiferencial respecto de uuuln\lnlnNatural logarithmConditionsln∣a+buu∣+C\ln\left|\frac{a + bu}{u}\right| + Clnua+bu+CFound an error? Contact us.Related formulasRelatedFormula∫u du(a+bu)2=ab2(a+bu)+1b2 ln∣a+bu∣+C\int \frac{u\,du}{(a + bu)^2} = \frac{a}{b^2(a + bu)} + \frac{1}{b^2}\,\ln|a + bu| + C∫(a+bu)2udu=b2(a+bu)a+b21ln∣a+bu∣+COpen formulaOpen in sectionRelatedFormula∫u2 du(a+bu)2=1b3(a+bu−a2a+bu−2a ln∣a+bu∣)+C\int \frac{u^2\,du}{(a + bu)^2} = \frac{1}{b^3}\left(a + bu - \frac{a^2}{a + bu} - 2a\,\ln|a + bu|\right) + C∫(a+bu)2u2du=b31(a+bu−a+bua2−2aln∣a+bu∣)+COpen formulaOpen in sectionRelatedFormula∫duu2(a+bu)=−1au+ba2 ln∣a+buu∣+C\int \frac{du}{u^2(a + bu)} = -\frac{1}{au} + \frac{b}{a^2}\,\ln\left|\frac{a + bu}{u}\right| + C∫u2(a+bu)du=−au1+a2blnua+bu+COpen formulaOpen in sectionRelatedA.10.2 With radicals a+bu\sqrt{a + bu}a+bu∫u a+bu du=215b2(3bu−2a)(a+bu)3/2+C\int u\,\sqrt{a + bu}\,du = \frac{2}{15b^2}(3bu - 2a)(a + bu)^{3/2} + C∫ua+budu=15b22(3bu−2a)(a+bu)3/2+COpen formulaOpen in sectionRelatedFormula∫duu(a+bu)=1a ln∣ua+bu∣+C\int \frac{du}{u(a + bu)} = \frac{1}{a}\,\ln\left|\frac{u}{a + bu}\right| + C∫u(a+bu)du=a1lna+buu+COpen formulaOpen in sectionRelatedFormula∫u dua+bu=23b2(bu−2a) a+bu+C\int \frac{u\,du}{\sqrt{a + bu}} = \frac{2}{3b^2}(bu - 2a)\,\sqrt{a + bu} + C∫a+buudu=3b22(bu−2a)a+bu+COpen formulaOpen in sectionRelatedFormula∫u2 dua+bu=12b3[(a+bu)2−4a(a+bu)+2a2 ln∣a+bu∣]+C\int \frac{u^2\,du}{a + bu} = \frac{1}{2b^3}\left[(a+bu)^2 - 4a(a+bu) + 2a^2\,\ln|a+bu|\right] + C∫a+buu2du=2b31[(a+bu)2−4a(a+bu)+2a2ln∣a+bu∣]+COpen formulaOpen in sectionRelatedFormula∫u2 dua+bu=215b3(8a2+3b2u2−4abu)a+bu+C\int \frac{u^2\,du}{\sqrt{a + bu}} = \frac{2}{15b^3}\left(8a^2 + 3b^2 u^2 - 4abu\right)\sqrt{a + bu} + C∫a+buu2du=15b32(8a2+3b2u2−4abu)a+bu+COpen formulaOpen in section